My goal in the next few posts is to talk about low order finite groups – that is, groups which contain a small number of elements.

My introduction to groups is going to be rather nonstandard. And it will be sketchy. Grab your favorite Introduction to Abstract Algebra or Introduction to Group Theory book. Suggestions:

- Fraleigh, “A First Course in Abstract Algebra”. A popular introductory text. I own two different editions.
- Dean, “Classical Abstract Algebra”, ISBN 0060416017. An excellent, if little known, introductory text.
- Dummit & Foote, “Abstract Algebra”. Written for undergraduates, with enough material for grad students. This was my main reading this time around.
- Schaum’s Outline of Group Theory. Cheap.
- Armstrong, “Groups and Symmetry”, ISBN 0387966757. An excellent undergraduate text with emphasis on the actions of groups on geometric figures. I’ve been using this book, too.

I think the cleanest starting point is to define a group as follows. We start with a set G, and a binary operation * on it. That is, given any two elements a, b of the set G, we have the product a*b, and that product is an element of G. Specifically, we require that the set be closed under the operation (or product). (That rules out, for example, the dot product of two vectors – because the dot product is not a vector, but a scalar.)

Second, we require that the operation * be associative. (This rules out, for example, the vector cross product – which is not associative, although the cross product of two vectors is again a vector.)

Third, we require that there exist an identity element e in the set G. That is, e satisfies two equations: for every element a in G, multiplying a by the identity – on the left or on the right – gives us a.

a * e = e * a = a.

Fourth, we assume that for every element a in G, there exists an inverse element (denoted ) such that the product of an element and its inverse is the identity:

.

Now is a good time to mention that

e * e = e

implies that the identity e is its own inverse.

One nontrivial example of a group – but it has an infinite number of elements – is the set of invertible nxn matrices (for fixed n), with matrix multiplication as the binary operation. I think this is a nice example because the operation is neither addition nor multiplication, but a combination of them.

Suppose n = 2. Then we’re speaking of invertible 2×2 matrices. By assumption all our matrices are invertible (that’s one requirement). We have an identity (that’s two):

Matrix multiplication is associative (that’s three). The product of two invertible 2×2 matrices is another invertible 2×2 matrix (and that’s four, the closure requirement). How do we know that? Because we have a formula for the inverse of a product:

.

That is, if A and B are invertible, we know that AB is invertible – because we’ve got its inverse in our grubby little hands; it’s .

Although not all 2×2 matrices are invertible, we have restricted our set G to be all the invertible ones and nothing else.

Note that matrix multiplication is not commutative: in general, AB ≠ BA.

Another example – a major one – is the set of integers mod n with addition as our operation. This is a finite group. Consider, for example, the integers mod 2. We have a set of two elements, often written 0 and 1. Any even integer is replaced by 0, and any odd integer is replaced by 1. We have

0 + 0 = 0

1 + 0 = 1 = 0 + 1

1 + 1 = 0.

We have closure, associativity, an identity (0), and that last line says that 1 is its own inverse. The second line shows that this group is commutative. Such a group is called abelian, in honor of Niels Henrik Abel (1802-1829… yes, he died young).

Now consider the integers mod 3. Our set of elements is often written {0, 1, 2}; then “mod 3” means that any positive integer a should be replaced by its remainder after division by 3. And we have

0 + 0 = 0

1 + 0 = 1 = 0 + 1

1 + 1 = 2

2 + 0 = 2 = 0 + 2

2 + 1 = 0 = 1 + 2.

Again we have closure, associativity, an identity (0), and that last line shows that 1 and 2 are inverses of each other. So -1 would be replaced by 2, and -2 would be replaced by 1.

It is not essential to use {0,1} or {0,1,2}. All of us should be familiar with the integers mod 12 – which we usually arrange in a circle, marked with numbers 1 thru 12 instead of 0 thru 11.

The groups of integers mod n are not only commutative – i.e. abelian – groups; they are also cyclic groups. That is, they can be generated by repeated operation on one element. For the integers mod 3, for example, we might start with 1 and keep adding 1:

1

1 + 1 = 2

2 + 1 = 0

and we just got all three elements. It turns out that every finite cyclic group is equivalent to the integers mod n for some n.

Having shown you a few examples, let me return to the theory.

There are a couple of things we can show immediately:

There is only one identity: e is unique.

For each element a in G, its inverse is unique. (That’s what let’s us write )

Then there are two more interesting things we can show.

We have left and right cancellation laws:

if a * b = a * c, then b = c

if b * a = c * a, then b = c.

We can find unique solutions to equations:

if a * x = b, then

if x * a = b, then .

Finally, we could have started with what look like less stringent requirements. Instead of assuming the existence of an identity – i.e. a two-sided identity e such that for all a in G

a * e = e * a = a,

we could have required only that there exist a right identity f such that

a * f = a,

and we could have required only that every element a have a right inverse A such that

a * A = f.

From those two one-sided definitions, we can show, however, that the right identity is in fact a left identity, and that the right inverse is in fact a left inverse – so we’ve come back to the original definition using “identity” and “inverse”.

That’s not just theoretical babble. Suppose I have a 3×3 matrix

After considerable effort by hand (imagine, no Mathematica… sob), I think that its inverse is

I’d like to check my answer, so I laboriously compute the product BA and I get BA =

Do I need to also compute AB? No: BA = I tells me that B is a left inverse of A – and I know therefore that B is also a right inverse, too, so I conclude that AB = I without computing it.

As it happens, I prefer to use a multiplicative notation for cyclic groups, and I prefer to see them as rotations of regular polygons.

Consider the cyclic group C2 with 2 elements: {1, g} such that g^2 = 1.

Geometrically, imagine an arrow… which we can rotate 180°. Do it twice and it’s back where it started.

We’ve already seen one realization of C2: the integers mod 2 under addition. We could also use {-1,1} under ordinary multiplication, since (-1)^2 = 1.

In general, the multiplication table for C2 = {1, g} looks like this:

(The entry in the table is xy; the group elements are 1, g.)

Now let me tell you where that picture came from. I am using a free package called “AbstractAlgebra”, which can be found here. I originally acquired it as a CD accompanying a book, “Exploring Abstract Algebra with Mathematica”, for Mathematica version 4 or 5. I am currently running a version for Mathematica version 7, and there is one available for version 8.

I find it useful sometimes, and very frustrating sometimes. Generating multiplication tables is useful. That I cannot get a permutation representation of a cyclic group is frustrating.

The web page says I can get an nth root representation, but I haven’t figured out how.

What about the cyclic group of order 3, i.e. with 3 elements. We expect {1, g, g^2}, with g^3 = 1. (We say that g is of order 3; so is g^2.) Do we have an inverse for g? Yes: g^2.

We could also realize C3 as the cube roots of 1:

If we let

then

(By the way, it’s not an accident that looks like a CCW rotation through .)

Finally, we could imagine C3 as the rotations of an equilateral triangle that leave it in what appears to be the same position (unless we mark the vertices). That is, if we have an equilateral triangle like so…

Then I can construct permutations which change the labels on the vertices… but I think of them as actually rotating the triangle.

Here’s the identity, in various forms:

Here’s the permutation that looks like a 120° clockwise rotation:

And here’s a 120° CCW rotation – or, equivalently, a 240° CW rotation:

Rotate thru 120° CW, and we get this…

Do it again, and we get this…

Do it a third time and we back to the beginning: we effectively did nothing.

So, we’ve seen multiple representations of C3 – permutations, cube roots of 1, rotations of an equilateral triangle – but they’re all essentially the same: they have the same Cayley Table. It turns out that there is “really” only one cyclic group of order n (the technical phrasing is, “up to isomorphism”).

What about a cyclic group of order 4? First, realize that we could take the four fourth roots of 1: among other things, this means that we know there exists a cyclic group of order 4. Not only do we have uniqueness… we also have existence.

Let’s see: (-1)^2 = 1 so -1 is of order 2… but i^2 = -1, so i and -i are of order 4. What’s the full multiplication table look like?

Without practice, that may not be very enlightening. But, for example, the three 1’s in the southeast corner show that g and g^3 are inverses and g^2 is its own inverse.

Let me construct the set of permutations that makes up C4. Here’s the identity.

A 180° rotation:

We see (you might have to look) that g and g^3 are of order 4; g^2 is of order 2. So we could match things up – “construct an isomorphism” between the permutations and the 4th roots of 1 – by taking g = i or g = -i.

Maybe it’s easier to just write {1,g,g^2,g^3} with g^4 = 1 and compute the orders of g^2 and g^3.

Geometrically (of course?) we draw a square and imagine rotating it thru 90° once, twice, thrice, and four times.

This progression, from equilateral triangle to square, goes on forever, for any finite n: we can take n distinct nth roots of 1, or we can image rotating a regular n-gon thru multiples of 360°/n, or we can take g^n = 1 with all lower powers distinct: {1, g, g^2, … , g^(n-1) }.

Having shown you cyclic groups, I’ll do dihedral, symmetric, alternating, and the quaternion group … in the next group theory post.

February 25, 2012 at 10:48 pm

Is it possible To create an S group out of the quaternion group. I realize that to create the group in its entirety would be next to impossible by hand, however I really only need to be able to show an isomorphism between the original quaternion group and a subgroup of the the permutations created from it. I have look everywhere and I have found books and articles that say it isn’t possible and those that say you can, but none show me how to do it with out spending an enormous amount of time doing so. My end result is to show that Cayley’s theorem. Any help would be awesome.

February 26, 2012 at 9:32 am

Look at the end of the post about the quaternion group; the last picture shows its Cayley table. And that shows you the 8 permutations that are isomorphic to the quaternion group. That is, you get an 8-element subgroup of which is isomorphic to the quaternion group.