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Theory: Group Theory

We already covered a great deal about groups in the chapter Theory: the structure of Abelian groups.  But well, those were Abelian.  What about non-Abelian groups?

About the most general way to think about algebraic systems is by thinking about matrices.  A set of NxM matrices filled with real numbers covers about anything except pure noise (at least, in my opinion).  Yes, real numbers, because also the complex numbers can be covered with it.  Lets do that as a first exercise.

Consider the set of matrices,

\[ F = \left\{ \begin{pmatrix} a & 0 \\ 0 & a \\ \end{pmatrix} \,\vert\, a \in \mathbb{R} \right\} \]

then this set is obviously isomorphic with $\mathbb{R}$ ($F \cong \mathbb{R}$) and therefore it is a field itself.  I will assume you are familiar with the rules of multiplying matrices, so check for yourself that this set of 2 $\times$ 2 matrices behaves exactly like ordinary real numbers.  Please note that the multiplicative identity is the matrix where a = 1, and the additive identity is the matrix where a = 0.  We will use that without further notice below, so don't get confused.

Next, lets extend the field F by adjoining a t to it:

\[ K = F(t) \]

Hey, I can't help it, that is the way to write that you extend a field with some new element and all elements that naturally follow from it in order to make the field closed again.  Note that adjoining an element to a ring is written R[t], using square brackets. We can also speak about the field extension K : F, and write the extension degree as m = [K : F].  Its just a new notation that I thought I'd throw in, we did this before.

However, in this case we don't specify a reduction polynomial for t but instead we specify exactly what it is:

\[ t = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \]

That doesn't mean that there isn't a reduction polynomial however.  It is easy to verify that

\[ t^2 + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

so the left hand side of that equation is the reduction polynomial (it is irreducible).  As you will have noticed, the square of t is equal to the additive inverse of the multiplicative identity:

\[ t^2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

And therefore we can identify t with the complex number i and establish that K = F(t) is isomorphic with $\mathbb{C}$.


Ok, back to arbitrary NxM matrices.  When the matrix isn't square (n $\neq$ m) then you can't multiply them.  You can still add them, but that isn't very interesting - that would be something isomorphic with a nm dimensional vectorspace and that is not what we want to investigate (we are not investigating matrices, we just use them as example), so forget about adding for now.  Of course you'd still be able to multiply a NxM matrix with a MxN matrix, but putting matrices of different shape into a set will never form a group because in a group (where the binary operator is multiplication) you need to be able to multiply any two elements.  In other words, we are only interested in square NxN matrices.

Actually, we are only interested in finite groups (so I guess the title of this chapter is a bit misleading).  That means we can also put a restrictions on the elements of the matrices: the elements of the matrices are finite (unit) rings (like $\mathbb{Z}_n$ for some n > 0 in $\mathbb{N}$).  A ring is set of elements for which addition, subtraction (additive inverse) exists and multiplication is defined and algebraically closed (the result of such an operation stays inside the ring) but not division because not all elements necessarily have a multiplicative inverse.  A unit ring is a ring with a multiplicative identity in it (1) and I like that because I insist to be able to construct identity matrix for our group that looks a bit familiar.

Another axiom of groups is that there must exist an inverse for every element.  Therefore we need to restrict ourself to matrices that have inverses.  Note that this means that a matrix with all zeroes is not an element of our group!  That's ok, we don't need it: we don't have an addition operator, so we don't need the additive identity (which is what zero is).

Karl Dahlke owns a wonderful site, a must for your bookmarks, with a large collection of lectures on many mathematical topics.  Below you will find a copy of his chapters on Group Theory, reproduced almost literally here with his permission.  Then each paragraph is followed by a block of text from my hand, clearly marked with a different background color.  My main goal with those extra sections is to provide (additional) examples, and sometimes clarifications.

A group

Copyright © 2002-2004 Karl Dahlke
Let S be the finite set of NxN matrices, with elements from a given finite ring, all of which matrices have an inverse.  Let x be one of those matrices.  Then xn eventually repeats an earlier value, say xm.  If m were 2, or anything larger, we would cancel an x from each side, so assume m = 1.  Thus xn = x.

Let e (the identity) = xn-1.  For any y, yx = yxn = yex, and by cancellation (x must have an inverse), yxx-1 = y = yexx-1 = ye.  Do the same on the left to show y = ey, hence e is an identity element.  There can only be one such element e, for if there were two, ef = e and ef = f, whence e = f.

Recall that every x in S leads to e eventually.  If xn-1 = e, then xn-2 is the inverse of x.  This breaks down if n = 2, but in that case xx = x, and x is in fact the identity element.  So we have found an identity and inverses, and the set becomes a group.

Example:

Let all the elements of the matrices be element of $\mathbb{Z}_3$ = {0,1,2}, and let the matrices be 2x2 matrices, then the set

\[ G = \left\{ \begin{array}{ccccccccc} \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 0\\ 0 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 0\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 0\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 1 & 2 \end{pmatrix}, \\ \begin{pmatrix} 1 & 0\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 0\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 0 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 0 & 2 \end{pmatrix}, \\ \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 0 & 1\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 0 & 1\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 0 & 1\\ 2 & 0 \end{pmatrix}, \\ \begin{pmatrix} 0 & 1\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 0 & 1\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 0 & 2 \end{pmatrix}, \\ \begin{pmatrix} 2 & 2\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 2\\ 0 & 2 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 2\\ 1 & 0 \end{pmatrix}, \\ \begin{pmatrix} 2 & 2\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 2\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 2 & 2\\ 2 & 1 \end{pmatrix} \end{array} \right\} \]

is a non-Abelian group under the operation of multiplication, with cardinality 48.

The order

Copyright © 2002-2004 Karl Dahlke
The order of x is n if n is the smallest positive integer satisfying xn = e.  Sometimes the notation |x| is used for the order of x.

An involution is an element with order two (its own inverse).

The order of the group g, written |g|, is the size of g, i.e. the cardinality of its underlying set.  A finite group has finite order.  If x generates the entire group, then |x| = |g|.

Example:

The matrix $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ has order 1.

The matrix $\begin{pmatrix} 1 & 0 \\ 2 & 2 \end{pmatrix}$ has order 2.

The matrix $\begin{pmatrix} 2 & 2 \\ 1 & 0 \end{pmatrix}$ has order 3.

The matrix $\begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix}$ has order 4.

The matrix $\begin{pmatrix} 2 & 2 \\ 0 & 2 \end{pmatrix}$ has order 6.

The matrix $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ has order 8.

All other 42 matrices also have either an order 2, 3, 4, 6 or 8.  There is one matrix with order 1 (the identity).  There are 13 matrices that are an involution, having order 2.  There are 8 matrices with order 3, 6 matrixes with order 4, 8 matrices with order 6 and 12 matrices with order 8.  This is not like the structure of any Abelian group.

Subgroups

Copyright © 2002-2004 Karl Dahlke
Let g be a group and let h be a subset of the elements of g.  Now h is a subgroup of g if h is a group, in and of itself, using the same group operator as g.  Example: for any positive n, the multiples of n form a subgroup of the integers.  If n > 1, the subgroup is a proper subgroup.

If h is a nonempty subset of the group g, h is a subgroup iff every a and b in h implies a/b in h.  The forward direction is obvious, so assume the latter.  Since h contains a/a, it includes the identity.  Since h includes 1/a, we have inverses.  Finally h contains ab, via a/(1/b), hence the operator does not map h to anything outside of h and is thus algebraically closed.

The following sets are examples of subgroups of G which are generated by the first element (subsequent elements are a power of the first element):

\[ H_1 = \left\{ \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\} \]

\[ H_2 = \left\{ \begin{pmatrix} 0 & 1 \\ 2 & 2 \end{pmatrix}, \begin{pmatrix} 2 & 2 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\} \]

\[ H_3 = \left\{ \begin{pmatrix} 0 & 1 \\ 2 & 1 \end{pmatrix}, \begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\} \]

Of course when P has order 6, so that P6 = e, then P2 will have order 3, and P3 will have order 2.  So that the following sets are also groups.

\[ H_4 = \left\{ \begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\} \]

\[ H_5 = \left\{ \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\} \]

These are just like the cyclic groups, order n, that we saw before when examining the Abelian groups.  They even commute!  After all, $P^x \cdot P^y = P^{(x + y \mod n)} = P^y \cdot P^x$.  Therefore, these cyclic groups are Abelian and isomorphic with Cn.

There are however also subgroups in G that, like G itself, are non-Abelian.  For example if we combine the elements of H1 with H4:

\[ H_6 = \left\{ \begin{array}{ccc} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 0 & 2 \\ 1 & 2 \end{pmatrix}, \\ \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix} \end{array} \right\} \]

And when now you expect the two new points to have order 6, then you lose.  Both new points are mysteriously an involution as well.  In fact, the largest cyclic subgroups they are a part of have order 2, so they are indistinguisable from the generator of H1.  Note that there are really not more elements in this group; if you multiply any two elements a, b $\in$ H6 either as ab or as ba then the result is again in H6.

Please be warned, these groups are by far not large enough to get an insight in the structure of non-Abelian groups, but it suffices to get an understanding of the concept subgroup.

Our example group G has 54 different subgroups.  We already mentioned the 28 cyclic subgroups above of rank 1.  Furthermore it has 26 subgroups of rank 2 (which means that there are a minimum of two generators needed to generate all elements).  G itself also has a rank of 2.  Then there is one subgroup of rank 2 with 24 elements, 3 subgroups of rank 2 with 16 elements, 4 subgroups of rank 2 with 12 elements each, 4 subgroups of rank 2 (and 3 of rank 1) with 8 elements, 8 subgroups of rank 2 (and 4 of rank 1) with 6 elements and 6 subgroups of rank 2 (and 3 with rank 1) with 4 elements.  The program that was used to brute force check this can be found in testsuite/point/groups.cc.

Cosets

Copyright © 2002-2004 Karl Dahlke
If you haven't seen cosets before, group theory can be a difficult place to start.  (It's so abstract).  Let me explain the idea of a coset in more general terms.

Let h be a substructure inside g.  Shifted copies of h are called cosets of h, and they cover all of g without overlap.  For example, let g be the xy plane and let h be the x axis.  For every real number c, the line running parallel to the x axis, through y = c, is a coset of the x axis.  These cosets cover the entire plane without overlap.

The quotient g/h is the collection of cosets.  In this example, the quotient is really (isomorphic with) the y axis.

With this in mind, let's build cosets of the subgroup h inside the group g.

Given a subgroup h, the right coset of an element x is the set of elements h$\cdot$x.  Left cosets are defined similarly, i.e. x$\cdot$h.

If a and b are in h, and ax = bx, then cancel x, and a = b.  All the elements in a coset are distinct.  there are |h| of them.

In fact the elements of any given coset, left or right, correspond one-on-one with the elements of h.

Thanks to the identity in h, x is always in the coset of x (reflexivity).  Since we can always multiply by the inverse of an element in h, x is in the coset of y iff y is in the coset of x (symmetry).  If z is in the coset of y is in the coset of x, z is in the coset of x (transitivity).  In other words, y = ax and z = by, hence z = bax.  We have an equivalence relation, and g can be partitioned into well defined cosets.  These cosets have the same size, or cardinality, namely |h|.

The above means that if you have some a that is an element of the subgroup h of g, then multiplying a with any element x of g causes the result to be part of a coset of h.  When x is also in h, then the result is also again in h of course, and the coset is really just h itself.  It gets more interestingly when you multiply with some x that is not in h.  The result will be that you get a complete copy of h, of the same size, none of whose elements are in h.  Moreover, you could have multiplied with any element of this coset and you'd have gotten the same coset: each element of the coset is thus a representative for that coset.  It follows also trivially that there are in total |g| / |h| cosets of h.

For example, let us multiply each element of H6 with some x not in H6.  Without changing the order in the coset, this will be the element that becomes below the identity then (reflexivity).

\[ H_6 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 2 \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}, \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix} \right\} = H_6 \]

\[ H_6 \cdot \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \left\{ \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 2 & 2 \end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 2 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix} \right\} = coset_1 \]

where the $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ was chosen arbitrarily, just something not in H6.

This was a right coset.  A left coset would have looked like this:

\[ \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \cdot H_6 = \left\{ \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}, \begin{pmatrix} 2 & 2 \\ 0 & 2 \end{pmatrix}, \begin{pmatrix} 2 & 2 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 2 & 2 \\ 2 & 1 \end{pmatrix} \right\} \]

Note that this does overlap (the first element) with the right coset.

If we had picked any other element of the right-coset coset1 that we found, it would have generated the same coset.  For example,

\[ H_6 \cdot \begin{pmatrix} 2 & 0 \\ 2 & 2 \end{pmatrix} = \left\{ \begin{pmatrix} 2 & 0 \\ 2 & 2 \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 2 & 2 \end{pmatrix} \right\} \]

which is indeed the same coset as coset1.

By picking yet another element of G, that is not in H6 nor in coset1, we can generate another right coset with 6 new elements, etcetera.  Since there are 48 elements in G and 6 elements in H6, we will be able to generate in total 8 distinct right-cosets from H6, each with 6 different elements, until all of G is covered.

Finally note that these cosets are not groups, they don't even contain the identity element.  However, cosets can act as group elements themselfs.  Consider a set S of cosets of a subgroup H of a group G.  Then we know that each coset has the same number of elements and that any element of G occurs only in one of the cosets.  We can write a coset of H represented by some x in G as:

\[ \left\{ xh \,\vert\, h \in H \right\} \]

or in short

\[ xH \]

As said before, any x element of this particular coset will lead to the same coset.

We can write the set of all cosets of H as:

\[ \left\{ xH \,\vert\, x \in G \right\} \]

Now let $A,B \in \left\{ xH \,\vert\, x \in G \right\}$.  In other words, A and B are cosets of H.

\[ A = \left\{ ah \,\vert\, h \in H \right\} = aH \]

and

\[ B = \left\{ bh \,\vert\, h \in H \right\} = bH \]

for some $a,b \in G$.  Then AB is defined as

\[ A \cdot B = \left\{ ah_1 \cdot bh_2 \,\vert\, h_1,h_2 \in H \right\} \]

When exactly S forms a group under this operation is decribed below in the paragraph Normal subgroup.

Lagrange's Theorem

Copyright © 2002-2004 Karl Dahlke
Lagrange's theorem is an immediate consequence of the above: |h| divides |g| whenever |g| is finite.  If |g| = 24, you don't have to look for a subgroup of size 7.  There isn't one.

Representative

Copyright © 2002-2004 Karl Dahlke
Sometimes we use designated elements from these cosets to represent them, i.e. coset representatives, somewhat like sending a representative from your district to congress.  For brevity, such a representative is called a cosrep.  As the name suggests, cosreps are usually taken directly from g, a true representative of the coset, but sometimes we rename the cosets, and assign them different symbols.  The group may have no numbers in it at all, but if the cosets behave like integers, we might rename them 1 2 3 $\ldots$ n.

We already saw that every element of a coset can represent that coset, as it will generate the same coset.  Now suppose we had a subgroup with 8 elements (we do), so that we could generate 6 distinct cosets from that (including the subgroup itself).  And further suppose that the 6 elements of H3 would be devided over those 8 cosets (where its identiy element would be taken from the subgroup of 8, of course).  Then we could use H3 as representatives of the cosets, and they would form a group isomorphic with $\mathbb{Z}$6, so we might as well rename those cosreps into 0, 1, 2, 3, 4, 5.  Unfortunately, I can't give an example of that because there is no such subgroup in G.  That is, it turns out that $\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$ is in every cyclic subgroup with order 6 and 8, so that aint gonna work; and I'm too lazy to come up with a totally new example at this point.

Normal Subgroup

Copyright © 2002-2004 Karl Dahlke
Let h be a subgroup of g.  If xh = hx for every x $\in$ g, then h is a normal subgroup.  In other words, its left and right cosets coincide.  This does not mean x commutes with the elements of h, it only means xh and hx produce the same set.  Of course, if h does commute with every x $\in$ g, then h has to be normal; every subgroup of an Abelian group is normal.

An equivalent definition says h is normal iff xy/x is in h for every x $\in$ g and every y $\in$ h.  If left and right cosets are the same, then xy = zx for some z $\in$ h, hence xy/x = z, a member of h.  Conversely, if xy/x is always some z $\in$ h, then xy is the same as zx, a member of the right coset.  We can run the other direction by replacing x with x inverse.  Multiply by x on the left and right, giving yx = xz.  Each member of the right coset is a member of the left coset, and h is normal.

A simple group has no normal subgroups, other than 1 and itself.  This may remind you of the definition of a prime number.  If g has no proper subgroups at all, it has no normal subgroups, and is simple.  $\mathbb{Z}$p, with p prime, is an example, having no subgroups by Lagrange's theorem.  An abelian group must be free of subgroups to be simple.

G has only one normal subgroup.

\[ N = \left\{ \begin{array}{cccccccc} \begin{pmatrix} 0 & 1\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 0 & 1\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 0 & 1\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 0\\ 1 & 1 \end{pmatrix}, \\ \begin{pmatrix} 1 & 0\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}, \\ \begin{pmatrix} 2 & 0\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 0 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 2 & 2\\ 0 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 2\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 2 & 2\\ 2 & 1 \end{pmatrix} \end{array} \right\} \]

Given that this is a subgroup of G, it is easy to see that it must be a normal subgroup: it has exactly half the elements of G.  That means that its left and right coset must coincide and thus it is a normal subgroup.

The set of cosets of N form a group iff N is a normal subgroup (of G).  This particular example of G and N aren't very instructive as the set of cosets of N only contains two elements: N and the set of all other elements of G.  Nevertheless, let us call that other set O, thus

\[ O = \left\{ \begin{array}{cccccccc} \begin{pmatrix} 1 & 0\\ 0 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 0\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 0\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 0\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 0 & 2 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 0 & 1 \end{pmatrix}, \\ \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 0 & 1\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 0 & 1\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 1 & 0 \end{pmatrix}, & \begin{pmatrix} 1 & 1\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 1\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 0 & 2 \end{pmatrix}, \\ \begin{pmatrix} 2 & 2\\ 0 & 1 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}, & \begin{pmatrix} 2 & 2\\ 1 & 2 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 2 & 1 \end{pmatrix}, & \begin{pmatrix} 0 & 2\\ 2 & 2 \end{pmatrix}, & \begin{pmatrix} 1 & 2\\ 2 & 0 \end{pmatrix}, & \begin{pmatrix} 2 & 2\\ 2 & 0 \end{pmatrix} \end{array} \right\} \]

And { N, O } acts as a group isomorphic with $\mathbb{Z}$2:

\[ \begin{matrix} N \cdot N = N \\ N \cdot O = O \\ O \cdot N = O \\ O \cdot O = N \end{matrix} \]

As you can see, N acts as the identity element here.  Let me make clear how this works by working out the details again like I did in the paragraph about cosets above.

Let o1 $\in$ O be a cosrep of O so that O = o1N.  Furthermore note that we can take any h $\in$ N be the cosrep of N, so also the identity matrix e: N = eN.  Then, for any coset A = aN of N we have that AN = aeN = aN = A.  And likewise NA = eaN = aN = A.  It isn't needed that N is a normal subgroup for that.

The only interesting case is therefore O2.

\[ O \cdot O = \left\{ o_1h_1 \cdot o_1h_2 \,\vert\, h_1,h_2 \in N \right\} \]

Because N is normal, the left and right cosets coincide and there will exist some o3 $\in$ O such that o1h2 = h2o3.  Realize that in general o1h2 $\neq$ h2o1!  And because h1h2 is again an element of N, lets say h, there will also exist some o2 $\in$ O such that h1h2o3 = ho3 = o2h and we see that

\[ O \cdot O = \left\{ o_1o_2h \,\vert\, h \in N \right\} \]

Now whatever coset of N this is, it is fixed of course, independent of the choice of o1 and o2.  It is easy to see that o1o2 cannot be an element of O.  Because, suppose that o1o2 = o $\in$ O, then there must exist a h $\in$ N such that o1h = o from which follows that o2 would be in N and it isn't.  Therefore o1o2 $\notin$ O.  Now, in our case, the only thing outside O is N, hence o1o2 $\in$ N for any two matrices o1 and o2 in O and thus $O \cdot O = N$.

More in general, we can derive that

\[ A \cdot B = \left\{ abh \,\vert\, h \in N \right\} = (ab)N \]

when N is a normal subgroup and a and b are cosreps of respectively A and B.  The parenthesis are redundant, but it shows that the result is again a coset of N, and thus that this operation on the cosets of N is closed.  But I am actually going on ahead of Karl Dahlke's lectures here.  Lets move on to his next paragraph.

Kernels and Quotients

Copyright © 2002-2004 Karl Dahlke
When the subgroup h is normal, the cosets form their own group.  The product of two cosets is found by multiplying any two cosreps together, and taking that coset.  We need to prove this is well defined, i.e. you always get the same coset no matter which cosreps you select.  If, instead of xy, we had selected xuyv, where u and v come from h, we replace uy with yw, since the left and right cosets of y are the same.  Now wv lives in h, so we have found another member in the same coset, the coset of xy.  We needed h to be normal, to replace uy with yw.

The identity represents the identity coset, and a coset's inverse is found by inverting one of its cosreps.  The resulting group of cosets is called the quotient group, or the factor group, and h is called the kernel.  In fact, we will now switch notation, whence k is the kernel and h is the quotient group.  Hope this isn't too confusing.

You'll notice that the quotient retains the coarse structure of the group, without all the fine detail; like looking at an object through the wrong end of a telescope.  $\mathbb{Z}$15 has a normal subgroup of {0, 5, 10}, and a quotient group with cosreps 0, 1, 2, 3 and 4, i.e. $\mathbb{Z}$5.  The quotient tells us that the original group has some similarity to the integers mod 5; the kernel tells us the group has something to do with $\mathbb{Z}$3.  Of course we could have used {0, 3, 6, 9, 12} as kernel, making $\mathbb{Z}$3 the quotient.  We don't always have the option to switch kernel and quotient, but in this case we do.

If g consists of the powers of a generator x, g is cyclic.  The only cyclic groups are the integers and the integers mod n, for all positive values of n.  The finite cyclic groups are denoted $\mathbb{Z}$n.  These groups are simple iff n is prime.  If n = pq, g can have the factor group $\mathbb{Z}$p or $\mathbb{Z}$q, as demonstrated by $\mathbb{Z}$15 above.

The above repeats more or less what I've already said before, except that it introduces two new names: the normal subgroup is called kernel (from now on denoted with K) and the set of cosets of this kernel, which form a group by themselfs thus, is called quotient.  This quotient is also be denoted with G/K in literature.  That notation is slightly counter intuitive as it gives one the feeling that the result is something smaller than G, and one wouldn't expect K itself to be an element of this quotient either.  But in fact, the quotient group G/K is just the set of all cosets of K within G, including the identity coset K itself.  And all cosets together still cover all of G.  The number of elements in G/K however is exactly equal to |G/K| = |G|/|K|, which makes the use of the slash a bit more logical.

Another notation found in literature for the cyclic groups $\mathbb{Z}$n (the integers ($\mathbb{Z}$) modulo n) is $\mathbb{Z}/$n.  Which is in fact a quotient group.  Therefore, in this case we have G = $\mathbb{Z}$ and we expect n to be some normal subgroup of $\mathbb{Z}$.  In fact, n is just a cosrep; and the subgroup of $\mathbb{Z}$ that it is refering to is n$\mathbb{Z}$ which are all integers divisable by n.  This is trivially a normal subgroup because $\mathbb{Z}$ is Abelian.  Of course, one also sees the notation $\mathbb{Z}/$n$\mathbb{Z}$ frequently used in literature.  This quotient then exists of all cosets of n$\mathbb{Z}$,

\[ \mathbb{Z}/n\mathbb{Z} = \left\{ x + n\mathbb{Z} \,\vert\, x \in \mathbb{Z} \right\} \]

Compare with

\[ G/N = \left\{ x \cdot N \,\vert\, x \in G \right\} \]

Please note that while we use multiplication for the matrices, and for the general theory in Karls lectures, the group operator of the integers $\mathbb{Z}$ is addition!  The cosets in {x + n$\mathbb{Z}$ | x $\in$ $\mathbb{Z}$} have a natural set of cosreps, namely those values of x for which 0 $\leq$ x < n.  Moreover, those cosets form a group that is isomorphic with $\mathbb{Z}$n.  They're thus not the exact same thing.

There is another nice example that gives insight.  Again this example is Abelian, unfortunately, but still worth remembering.  Consider vectors in a 2D plane through the origin.  They form a group under the operation of addition.  The origin is the identity element, inverses are vectors with the same length but pointing in opposite direction and adding any two vectors in the plane will result again in a vector in that same plane, etc.  Now, in 3D space, take a vector x that is not an element of this plane, but instead an element of the rest of the 3D space (this 3D space is our G, which obviously also a group).  The plane through the origin (K) is thus a (normal) subgroup of the 3D space (it is trivially normal because this is Abelian).  Adding the vector x to the plane results in a plane not going through the origin, but parallel to the orignal plane.  This new plane, x + K, is a coset of the original plane K.  The complete set of cosets of K, G/K, fills all of the 3D space, all of G.  All those planes together form a group that are isomorphic with a group of just a set of cosreps of those planes, for example, a line perpendicular to the planes (although any straight line would do).  This line is indeed again a group, as G/K should be.  Moreover, in this case we can, as was explained in the example of Karl too, exchange the kernel and the quotient: we can take any line as kernel (because, since this is Abelian, also the line is a normal subgroup) and observe that the quotient with this line is isomorphic with a plane through the origin.

Homomorphisms and Isomorphisms

Copyright © 2002-2004 Karl Dahlke
Taken from an old Greek word for shape, the various "morphisms" preserve the structure of a group, or at least part of its structure.  Actually these terms are applied to many different objects, not just groups.  If an object is based on a set, with one or more operators that define that object, and a function f maps this set to another set, and f preserves the defining operators, then f is a homomorphism.  Rings and fields have two operators that must commute with f; groups have only one.  Let's look at a simple example.

Let $\mathbb{Z}$ be the group of integers under addition and let f map $\mathbb{Z}$ onto $\mathbb{Z}$ mod 7.  Note that we can take x+y mod 7, and that's the same as taking x mod 7 and adding it to y mod 7 (and then again doing the result mod 7).  In other words, f(x + y) = f(x) + f(y).  That's it; that's all we need for a group homomorphism.

Every normal subgroup defines a group homomorphism.  If k is a normal subgroup, and h the resulting quotient group, let the function f take an element x $\in$ g to the coset of k represented by x ($f: x \mapsto xk$).  As described in the previous section, f is a function from g onto h, where g and h are both groups; but is f a homomorphism?  Does f commute with the multiplication operator?  Well the coset of x times the coset of y is, by its very definition, the coset of xy.  Thus f is a group homomorphism.

Conversely, suppose f is a homomorphism from the group g onto the set h, which possesses a preexisting operator $\times$.  Since f is a homomorphism, f(x)$\times$f(y) = f(x$\times$y).  In other words, f commutes with $\times$.  It is not hard to show that h is a group, inheriting all the group properties through f.  If g were a ring, h would be a ring, and so on.

Let 1 (the identity) in g map to an element we will call 1 in h, and verify that 1 acts as an identity element in h.  That is, f(1)f(x) = f(1x) = f(x), hence 1 leaves h where it is.

Let k be the set of elements in g that map to 1 in h.  For any x and y in k, f(x) = f(y) = 1 in h.  Their product is 1 in h, hence f(xy) = 1, hence xy is still in k.  Inverses must also map to 1 in h, and are present in k, hence k is a subgroup of g.

It is easy to show that xk/x remains in k, as f(xk/x) = 1, hence k is a normal subgroup of g.

Finally, the elements of h correspond to the cosets of k, and are multiplied acccordingly.  The image group h is indistinguishable from the quotient group g/k.  A group homomorphism defines, and is defined by, a normal subgroup and quotient group.

If the kernel of a homomorphism is the identity element, f is a monomorphism.  This comes from the Greek word mono, meaning one; the map is one-on-one.

If f carries g onto the entire group h, rather than a piece of h, f is an epimorphism.  This is usually what we mean when we say f maps g to h.  Sometimes we say f maps g "onto" h - same thing.  If f maps g "into" h, it is not always an epimorphism.  Take the integers mod 7, then multiply by 2.  This maps $\mathbb{Z}$ onto the even numbers mod 14.  It is a group homomorphism into $\mathbb{Z}$14, but it is not an epimorphism, because nothing maps to the odd numbers.

If f is an epimorphism and a monomorphism it is an isomorphism.  Elements map one-on-one, and all of g maps to all of h.  The map can be reversed to produce an inverse homomorphism.  The two groups are indistinguishable.  We have merely assigned different labels to the elements of g.  Some people would say these are in fact the same group; where the structure defines the group, without regard to the underlying set.  This abstract definition is valid, because group isomorphism is an equivalence relation.  We can clump all the isomorphic groups together and declare them a single group.

For example, if p is prime there is only one group (structure) of order p, namely $\mathbb{Z}$p.  Let x be an element, other than 1, of some group g with p elements, and consider the powers of x.  If xn becomes 1 before n reaches p, then x generates a proper subgroup, which is forbidden by Lagrange's theorem.  Hence g is merely the powers of x, and the exponents form the integers mod p.  After some judicious relabeling, the group is $\mathbb{Z}$p.

Endomorphisms and Automorphisms

Copyright © 2002-2004 Karl Dahlke
An endomorphism is a homomorphism that maps g into itself.  Multiply the integers by 29, and you have an endomorphism from $\mathbb{Z}$ into $\mathbb{Z}$.  It also happens to be a monomorphism, a perfect copy of $\mathbb{Z}$ inside itself.

An automorphism is an endomorphism, a monomorphism, and an epimorphism.  In other words, a function f maps g onto itself, and is one on one, and preserves the structure of g.  Multiply the integers by -1 for an automorphism.

To illustrate, lets look at the automorphisms of the group $\mathbb{Z}$n.  This is a cyclic group, generated by 1, so once f(1) is defined, the entire group is mapped.  The resulting map is an automorphism iff f(1) is coprime to n.  There are $\phi$(n) such automorphisms, including the trivial automorphism that maps 1 onto 1.  The integers have only two automorphisms; mapping 1 to $\pm{1}$.

The set of automorphisms forms a group.  Two functions are "multiplied" by applying the first, then the second.  These functions are really permutations on the underlying set, that happen to commute with the group operator.  Applying permutations is associative, and the identity permutation is the identity automorphism, we only need to prove that inverse automorphisms exist.

An automorphism f has an inverse permutation j; just move the elements back into position.  What if we apply j first?  Let z = j(x)*j(y).  Now f(z) = f(j(x)*j(y)) = f(j(x))*f(j(y)) = x*y.  Like f, j commutes with *, and is an automorphism.  The automorphisms form a group under function composition.

Return to $\mathbb{Z}$n, with its $\phi$(n) automorphisms.  If the first function maps 1 to u, and the second maps 1 to v, their composition maps 1 to uv.  The group of automorphisms is $\mathbb{Z}$n*, the elements coprime to n under modular multiplication.  Click here for a complete characterization of these multiplicative groups.

Inner Automorphisms

Copyright © 2002-2004 Karl Dahlke
The conjugate of an element w with respect to x is x*w/x.  If w and x commute then the conjugate of w is w

Given a group g and an element x, replace every element w in g with its conjugate. In other words, f(w) = xw/x.  This map commutes with *, and is an automorphism.  Specifically, it is an inner automorphism.  If g is abelian, xw/x = w, and every inner automorphism becomes the trivial automorphism.

An automorphism that is not an inner automorphism is an outer automorphism.

The composition of two inner automorphisms, derived from x and y, is the inner automorphism derived from yx.  One can use x inverse to produce the inverse of a specific inner automorphism.  Thus the inner automorphisms form a subgroup of all the automorphisms of g.

If x produces an inner automorphism, and f is any other automorphism, having inverse j, apply f, then the x automorphism, then j.  If w is a group element, we have j(x*f(w)/x).  Since j commutes with *, this is j(x)*j(f(w))/j(x), or j(x)*w/j(x).  The composition gives another inner automorphism, hence the subgroup of inner automorphisms is normal inside the (possibly larger) group of automorphisms.  This implies a quotient group, the automorphisms mod the inner automorphisms.

The center of g are those elements that commute with all of g.  They produce trivial inner automorphisms; if x commutes with everything then xg/x leaves g fixed.  Conversely, if x is not in the center of g then xy/x is different from y for some y, and the resulting inner automorphism is nontrivial.

Map g onto the inner automorphisms of g by carrying x to the function xg/x.  We must show that this is a homomorphism.  Compose xg/x with yg/y and get yxg/x/y, the inner automorphism associated with yx.  Technically, we were hoping for the inner automorphism associated with xy.  This is not a show stopper though.  Change the convention, so that the composition of two automorphisms is the second, followed by the first, rather than the first followed by the second.  The automorphisms of g still form a group.  Now the composition of xg/x and yg/y is xyg/y/x, as it should be, and the map from g onto its inner automorphisms is indeed a group homomorphism.

We showed that the kernel of this map is precisely the center of g.  Thus the inner automorphisms of g, with function composition defined as above, form a group that is isomorphic to g mod its center.

Direct Product, Direct Sum

Copyright © 2002-2005 Karl Dahlke

Let g and h be arbitrary groups, and let j be a larger group whose elements, as a set, are defined by the cross product of g and h.  Let g1 and g2 be members of g, while h1 and h2 are members of h.  The composite group j is the direct product of g and h, if the product of any two elements g1,h1 $\cdot$ g2,h2 in j is equal to g1g2,h1h2.  In other words, the groups g and h run in parallel.

We've already seen examples of this.  The group $\mathbb{Z}$7$\times$$\mathbb{Z}$7 contains 49 elements, two parallel copies of the integers mod 7.  For instance: 2,5 + 3,4 = 5,2.

Project j onto g or h by throwing away the "other" component.  This is a group homomorphism, hence either g or h can act as the kernel of j, with h or g playing the role of factor group.

Apply the above several times to take the direct product of a finite number of groups.

The direct product of an infinite set of groups is also well defined.  We don't have to worry about the axiom of choice, because every set has a special element, the group identity.  The direct product includes the element 1,1,1,1,1,1... which acts as the identity element for the composite group.  Group operations are performed per component.

The direct sum is essentially defined in the same way as the direct product: group operations are still performed per component, but almost all components are set to the identity element.  See more on the direct sum here.

As an example, consider the direct product or sum of $\mathbb{Z}$p for all primes p.  The prime cycles run in parallel, and independent of each other.  The element 0,0,0,0... is the group identity.  In the direct product, the element 1,2,3,4,5,6,7,8,9... is well defined.  In the direct sum, almost all components must be 0, as in 1,2,3,0,5,0,0,0,0... zeros thereafter.

Semidirect Product

Copyright © 2002-2004 Karl Dahlke

The semidirect product is a bit trickier.  Let g and h be arbitrary groups as above, and let a group homomorphism map h into the automorphisms of g.  In other words, every element x in h is associated with an automorphism on g.  Call this automorphism fx.

Use the "group action" convention, where the automorphism defined by xy is the automorphism of y, followed by the automorphism of x.  This seems backwards, but it's actually more convenient.  Naturally, f1 = 1, the trivial automorphism, and the automorphism f1/x is the inverse automorphism of fx.

If automorphisms have been assigned to the generators of h, then we have the entire map from h into the automorphisms of g.  If h is cyclic, for instance, generated by x, we only need know fx.  Then fx*x is fx applied twice, and so on.

Now lets define the semidirect product.  The product of two elements g1,h1 $\cdot$ g2,h2 has the following formula.

g1*fh1(g2), h1h2

Is j a group? Well 1,1 is certainly the two-sided identity.  Let's bring in g3,h3 and verify associativity. The second component, from the group h, is clearly associative; we only need look at the first component.

First component of (j1 * j2) * j3 =

g1 fh1(g2) fh1h2(g3) =

g1 fh1(g2) fh1(fh2(g3)) =

g1 fh1(g2fh2(g3)) =

first component of j1 * (j2 * j3).

Does g1,h1 have an inverse?  Let f be the automorphism associated with h1.  Apply f inverse to the inverse of g1 and call this g2.  Sure enough, g1f(g2) produces the identity in g, hence g2,h2 is the inverse of g1,h1, and j is a group.

A simple homomorphism maps j onto h, with g as kernel.  Thus j is the "semidirect product of g by h".

Necessary and Sufficient Conditions

Copyright © 2002-2004 Karl Dahlke

Let j be a group with kernel g and factor group h.  Thus j is an "extension of g by h".  But is j a semidirect product?

If it is, then j contains a copy of h.  Furthermore, the group h that lives inside j maps, by our group homomorphism, onto the factor group h.  Seen another way, we can select a cosrep for every coset of g in j, and these cosreps form a closed subgroup of j.  This subgroup is a mirror of h inside j.

Conversely, let h be a set of cosreps of g in j, while h is a closed subgroup inside j.  Of course h also represents and defines the factor group.

Let x be any element of h.  since g is the kernel, xg/x equals g, as a set.  That is, conjugation by x permutes the elements of g.  Apply this to g1g2, and the result is the same as xg1/x * xg2/x.  Therefore conjugation by x implements a group automorphism on g.  Call this automorphism fx.  This holds for every x in j, so it certainly holds for every x in h.

Since left and right cosets are the same, let the members of h represent right cosets.  The elements of j can now be labeled unambiguously.  Each element is gi,hi for some gi in g and some hi in h

Now consider the product g1,h1 * g2,h2.  Realize that x,y is the same as x*y, and you're almost there.

g1,h1 * g2,h2 =

g1 * h1 * g2 * h2 =

g1 * h1 * g2 / h1 * h1 * h2 =

g1 * fh1(g2) * h1 * h2 =

g1 fh1(g2), h1h2

The group j, with kernel g and factor group h, is a semidirect product iff a copy of h exists inside j, and maps onto the factor group h.  Such a group is called "split exact".

Consider the cyclic group of order p2 for some prime p.  Use the integers mod p2.  The multiples of p form the kernel, and the integers mod p form the factor group.  We have g and h in hand, but is Zp2 a semidirect product?  Let x be an integer that is 1 mod p, such that x generates a copy of h inside j.  The elements x, 2x, 3x all map to 1 2 3 etc in the factor group, but px is not 0.  The subgroup is not closed.  This is a group that has a kernel and quotient, but is not a semidirect product.

Relatively Prime Subgroups

Copyright © 2002-2004 Karl Dahlke

Let j be a finite group, with normal subgroup g, and subgroup h, such that |g| and |h| are relatively prime.  Let c and d be elements of h that happen to lie in the same coset of g.  Thus c/d also lies in g.  This means the order of c/d divides into |g| and |h|, which are relatively prime.  Thus c/d is the group identity, and c = d.  The members of h represent distinct cosets of g in j.  In other words, the members of h represent the factor group j/g.  Multiplication in h corresponds to multiplication in j/g, and h is in fact a copy of the factor group, living in j.  The group is split exact - a semidirect product.

Next suppose g is normal in j, and |g| and |j/g| are relatively prime, and j/g is known to be cyclic.  Let c be an element of g that generates j/g.  If |j/g| = u, cu is back in g, and has some order v in g.  Replace c with cv.  Since v and u are coprime, we are merely selecting a new generator for j/g.  Also, cu is the identity element in g, and in j.  Thus c generates a copy of the factor group inside j, and j is split exact - a semidirect product.

Joining and Intersecting Subgroups

Copyright © 2002-2004 Karl Dahlke

Using the a/b criterion, the intersection of arbitrarily many subgroups is a subgroup.  Using the x*h/x criterion, the arbitrary intersection of normal subgroups is normal.  We don't have to worry about an empty intersection; every subgroup contains the identity element.

The union of two normal subgroups may not even be a subgroup.  When multiples of 3 and 5 are combined, they don't contain the number 8, and are not a subgroup of $\mathbb{Z}$.  However, any set of elements, including the union of preexisting subgroups, can be used to generate a new subgroup. 

If q is a collection of group elements, let r contain q, the inverses of q, and all finite products taken from q and its inverses.  If a and b are in r, then a/b is in r, hence it is a subgroup, in fact the smallest subgroup that contains q.

Next expand r, so that it includes xw/x for every w in r and x in g.  This is still a subgroup, possibly larger.  Note that y(xw/x)/y = (yx)w/(yx) is in r, so r satisfies the criterion for a normal subgroup.  In fact r is the smallest normal subgroup containing q.

Joining a Subgroup and a Normal Subgroup

Copyright © 2002-2004 Karl Dahlke

Let k be normal in g and let h be any subgroup of g.  Let r be the subgroup generated by k and h, also called k join h, and let j be the intersection of k and h.  For every x in g, x*k/x winds up in k.  This still holds if x belongs to r, hence k is normal in r

Except for j itself, each left coset of j is completely separate from j.  If part of a coset of j is in h, the entire coset is in h, since all of j is in h - and similarly for k.  A coset of j might belong to k, or h, but not both, as j is their intersection.

Let x be the left cosets of j in h, and y the left cosets of j in k.  Formally, x is a collection of cosreps, somewhat arbitrary, and so is y.  Let 1, the group identity, represent j in both x and y.  Since j is the complete intersection of h and k, 1 is the only element common to x and y

If an element e1 is x1y1j1, and e2 is x2y2j2, what can we say about e3, the product of e1 and e2  It begins life as x1y1j1x2y2j2.  Since k is normal in r, the left and right cosets of x2 are the same.  We can move y1j1 to the right of x2, replacing it with a new value from k.  This gives x1x2y3j3y2j2.  Now everything to the right of x2 is an element of k, and x1x2 is another coset of h, so the element e3 is the product of elements taken from x, y, and j, in that order.  Since r consists of finite products from h and k, all of r is represented by these triples.

If two triples lead to the same element, i.e. x1y1j1 = x2y2j2, multiply by x2 inverse on the left.  Now the product x2 inverse times x1, call it x3, is multiplied by something in k, and produces something in k.  Clearly x3 is in k, and in h, hence in j, so the original cosreps x1 and x2 are the same.  Divide these out, leaving elements in k.

Now y1 = y2 and j1 = j2, and the triples were in fact the same.  Elements of r correspond one on one with triples drawn from x, y, and j.

If r is finite, the order of r is |h|*|k|/|j|.

Don't assume you can just multiply components together.  After all, r is not a direct product.  When we pull elements past each other, y1 is replaced with y3, and so on.  However, the projection onto h/j is a proper group homomorphism.  This is not surprising, since k is normal in r.  If e1 begins with x1, and e2 starts with x2, then e1e2 starts with x1x2.  The cosets of j in h faithfully represent the cosets of k in r.

Joining Disjoint Normal Subgroups

Copyright © 2002-2004 Karl Dahlke

Let h and k be normal in g, and let r be h join k, as described above.  Now the projection onto h/j and the projection onto k/j are both group homomorphisms.  In other words, we can simply multiply x1x2 and y1y2.

If h and k are disjoint, intersecting only in the identity element, then we don't have to worry about j at all.  Thus r becomes the direct product of h and k.

The Index of a Subgroup

Copyright © 2002-2004 Karl Dahlke

The index of a subgroup h in a group g is the cardinality of the left or right cosets of h.  This is well defined, at least for finite groups.  Lagrange's theorem tells us the index of h in g is |g|/|h|, whether we use left or right cosets.  This may not hold for infinite groups.  After all, $\infty$$\times$ 7 is the same as $\infty$$\times$ 9.

Index is always well defined when h commutes with g, or more generally, when h is normal in g, whence the index is the size of the quotient group g/h.

If k is a subgroup of h, each element of g lies in a coset of h, and after that cosrep is divided out, the remainder lies in a coset of k inside h.  Therefore the index of k in g is the index of k in h times the index of h in g.

If h has index 2 in g it is normal.  Choose any x not in h and observe that x*h \= h*x.  After all, there is only one coset outside of h.

Center, Centralizer, Normalizer

Copyright © 2002-2004 Karl Dahlke

The centralizer of a subgroup is the set of elements that commute with everything in the subgroup.  Verify that the centralizer of h is a group.  Apply the a/b test.  Inverse is the only tricky part.  If b commutes with h, does b inverse?  Given h/b, multiply by 1 = (1/b)*b on the left.  This simplifies to 1/b times h.

The center is the centralizer of the entire group.  In other words, the center is the subgroup that commutes with all of g.  The center is always a normal subgroup.

The normalizer of a subgroup h is the set of all elements x such that xh/x = h.  Clearly h is in the normalizer of h.  Remember that conjugation is an inner automorphism, one on one and onto.  When x comes from the normalizer of h, xh/x implements an automorphism on h.  This can be reversed, hence 1/x implements an automorphism on h, and 1/x is in the normalizer.  Similarly, xy is in the normalizer when x and y are both in the normalizer.  Thus the normalizer is a group.

Since everything in the normalizer fixes h by conjugation, h is normal in its normalizer.

Don't confuse the normalizer of h with the normal subgroup generated by h.  They are quite different.  When h is normal in g it is its own normal subgroup, yet its normalizer is all of g.  Conversely, the smallest normal subgroup containing h might be all of g, and the normalizer of h could simply be h.

Correspondence Theorems

Copyright © 2002-2004 Karl Dahlke

The correspondence theorem, or isomorphism theorem, is sometimes presented as three separate theorems.  In fact they are often called the first second and third isomorphism theorems.  Well I couldn't remember where to draw the line, so I just clumped them all together.  Hope that's ok.

This theorem, or theorems if you prefer, asserts the equivalence of the subgroups or normal subgroups of g with those in the factor group h.  In other words, subgroups containing k, or normal subgroups containing k, correspond one on one with subgroups or normal subgroups in the factor group g/k.  Let's get started.

Let k be a normal subgroup of g, with factor group h.  If r is a subgroup of g, its image s is a subgroup of h.  Conversely, if s is a subgroup of h, its preimage r is a subgroup of g.  Use the a/b criterion to verify this.

Applying the x*w/x criterion, verify r is normal in g iff its image is normal in h, and s is normal in h iff its preimage is normal in g.

Assume h is normal in g, and k is normal in both h and g.  Let q be the factor group g/k.  By the above, the image of h in q is a normal subgroup of q.  This leads to a factor group r.  We will show that r is the same as g/h.

The map is straightforward, though somewhat technical.  Given a coset of h in g, choose a cosrep x, let it define a coset of k in g, i.e. an element of q, and let that element of q define a coset of the image of h in q, i.e. an element of r.  We need to show this is well defined.  Multiply x by y for some y in h, and see if this changes anything.  Moving to q, the image of x is multiplied by something in the image of h.  This leads to the same element of r.  So it doesn't matter which cosrep we use.

Use the property of group homomorphism, from g to q to r, to show this map commutes with *, and is a group homomorphism.  Since q is the image of g and r is the image of q, the map is onto.  Finally, different cosets of h in g lead to different cosets of the image of h in q, hence the map is an isomorphism.

If g is finite, the index of k in g (|q|), is equal to the index of k in h (the size of the image of h in q) times the index of h in g (the size of r).

Dihedral and General Linear Groups

Copyright © 2002-2004 Karl Dahlke

I know you've been waiting for this: a non-abelian group that is easy to understand.

The dihedral group Dn is the rotations and reflections of a regular n-gon.  Picture an n-gon with thickness, similar to a coin with edges.  In fact dihedral comes from the latin, meaning two surfaces, like heads and tails.  Assign numbers to the edges of the coin, so you know what is going on.  Place the coin on the table with edge #1 facing front.  Rotate the coin one notch clockwise, then turn it over.  Edge #1 has been shifted to the left, then flipped over to the right.  Edge #2 is directly in front of you.  Now reverse the operations; flip the coin and then turn it clockwise.  Edge #1 still faces front after the flip, then shifts left.  This is not where it was before, hence the group is nonabelian.

The subgroup of pure rotations is Zn, and it has index 2, hence it is normal in the dihedral group.  All other group elements are involutions, except for the identity element (leave the coin alone).

For n = 1, the dihedral group is Z2, like a stick standing up, flipping end over end.

For n = 2, the coin has two edges, a fron edge and a back edge, and some thickness, so we can flip it over.  The group consists of vertical flips cross horizontal flips.  These are independent of each other, hence the group is Z2$\times$Z2.  This is not a cyclic group, not the same as Z4, but it is abelian.

When the coin has 3 or more edges, it looks like a traditional n-gon, and the group is non-abelian.

The general linear group of order n is the set of non-singular n$\times$n matrices under multiplication.  The special linear group is the kernel of the homomorphism implemented by the determinant, namely the n$\times$n matrices with determinant 1.  These groups are written GLn(r) and SLn(r), for n$\times$n matrices over a ring r.

Even and Odd Permutations

Copyright © 2002-2004 Karl Dahlke

A permutation of n elements is often written as an ordered list of the numbers 1 through n.  For instance, the permutation x = 23154 moves the element in position 1 to position 2, 2 to 3, and 3 back to 1, while elements in positions 4 and 5 are swapped.

If x and y are two such permutations, their composition moves i to y[x[i]].  Let y = 52134, and compose by writing y[x[i]] in sequence.  Hence xy = 21543.

Given a permutation, count how many times a larger number precedes a smaller number in the list.  If this count is even the permutation is even, else it is odd.  The identity permutation, 12345, has zero pairs out of order, hence it is even.  The permutation 54321 has ten pairs out of order, and is also even.

If two adjacent elements are swapped, i.e. a transposition, the parity of the permutation is reversed.  The swap only changes the pair in question; all other pairs are unaffected.  The swapped elements are moved into or out of ascending order, hence the parity is flipped.

Every permutation can be implemented by a series of transpositions.  There may be many ways to do this, but an even permutation is always achieved using an even number of transpositions, and an odd permutation requires an odd number of transpositions.

The composition of even permutations is even.  The first is achieved using an even number of transpositions, and so is the second, hence the composition is built using an even number of transpositions, and is even.  In general, even and odd permutations add together just like even and odd numbers.  Parity is a group homomorphism from the permutation group g into Z2.

If g includes odd permutations, the even permutations form a proper subgroup that maps to 0 under parity, while the odd permutations map to 1.  The even permutations form the kernel of the parity homomorphism, and are a normal subgroup in g.

Symmetric and Alternating Groups

Copyright © 2002-2004 Karl Dahlke

The symmetric group on n letters, written Sn, is the group of all possible permutations on n letters.  The order of Sn is n!.

The alternating group on n letters, written An, is the group of all even permutations on n letters.  The order of An is n!/2.  An is normal in Sn, in fact it is the kernel of the parity homomorphism.

The group An+1 defines the group of rotations of a generalized tetrahedron in n space, while Sn+1 defines the group of rotations and reflections.  This can be seen by placing any vertex in position, then the next, then the next, and so on, and reflecting if the last two must be swapped.

For $S_{\infty}$, parity, and hence $A_{\infty}$, may be well defined if permutations transpose only a finite number of letters in a linearly ordered set.  Once again $A_{\infty}$ is the kernel of the parity homomorphism, and is normal in $S_{\infty}$.

Coliniations

Copyright © 2002-2004 Karl Dahlke

A coliniation is a permutation group that carries subsets of points to other subsets of points.  The name coliniation is related to the word colinear, where the predefined subsets all contain two points.  Each permutation carries pairs of points to other pairs of points, hence lines to lines.  Each graph determines a coliniation.

Consider a regular n-gon.  A permutation begins by mapping one vertex onto any other vertex in the n-gon.  But lines must map to lines, so the adjacent vertex must map to a vertex that is adjacent to the image of the first.  This continues as we work our way around the n-gon.  The coliniation is Dn, as described earlier.

Copyright © 2002-2008 Carlo Wood.  All rights reserved.