Main Page   Reference Manual   Compound List   File List  

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 2x2 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 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.

Copyright © 2002-2004 Carlo Wood.  All rights reserved.