## Wednesday, April 26, 2006

### Groups, Subgroups, and Group Actions, Oh my! (REVISED)

Sorry for the bad title. While I was working on this, it popped into my head, and it won't go away. If I'm going to go crazy hearing that in my head, you can too.

Today's a bit of a definition day. To be able to really show how some of the stuff I've talked about informally works, you need to know some terminology. Don't worry - it's not entirely dull! (I'm also rather overtired this evening, so don't be surprised if some typos slip through; please let me know if you notice any.)

(Update: as the comment above indicated, I originally wrote this in a hurry while overtired. This was a very bad idea. As a result, it has been substantially rewritten to fix it up.)

### Subgroups

I've mentioned subgroups briefly once or twice; it's worth pulling it out and making it first class here.

A subgroup S of a group G is a group with the same group operator as G, and whose members are a subset of the members of G. Because the subgroup is also a group, that means that it needs to be closed under the group operation and inverse. To be precise:

A group S is a subgroup of G iff:
` members(S) subset members(G) all x,y in S : x*y in S all x in S : x^-1 in S`

There is a particular kind of subgroup that is interesting, called a normal subgroup. A subgroup S of a group G is a normal subgroup of G if/f:
` all x in G : (all y in S : x * y * x^-1 in S)`
Another way of writing that is:
`all x in G : x * S * x^-1 is a subset of S`
It happens to be provable for normal subgroups that the "is a subset of" up there can actually be replaced by "=".

A group is an Abelian Group if/f the group operator is commutative:
`all x,y in G : x * y = y * x`

If a group is abelian, then all of its subgroups are normal.

A trivial group is a group containing only the identity value.

A simple group is a group whose only normal subgroups are the trivial group and the group itself.

### Group actions

We're getting closer to being able to talk about the fun stuff. But we need to talk about group actions.

A group action is something that you can create by treating the members of the group as mappings. The way that you can do this is related to the symmetric groups.

The other day, when I talked about the symmetric groups. I defined the symmetric group of order N as a group over the integers from 1 to N. You can also define the symmetric groups in another way: the symmetric group over a set A is the set of all permutations of the members of the set. The symmetric group over a set A is written S_{A} The two are just different ways of expressing the same basic thing: the group of all permutations over a set of values.

Now, also remember that every group is isomorphic to a subgroup of some symmetry group. What this means is that given an appropriate set A, any group is equivalent to a permutation group over the set. Each member of the group is a function mapping A to A. The group action of a group G is a homomorphism from G to the symmetric group over A, S_{A}.

If you have a group G and a set A, then a group action of G on A is a function f such that:
• all g,h in G : (all a in A : f(g*h,a) = f(g,f(h,a)))
• all a in A : f(1,a) = a. (where 1 is the group identity element).
That's really a hairy way of saying:
1. There's a mapping from elements of the group to functions over the set;
2. Performing the group operation on the elements of set according to the mapping applies a symmetric transformation to the set.
Remember when I talked about the symmetries of a pentagon? Rotation, mirroring, etc. Well - for each of those symmetries, there is a group whose operation defines the particular kind of symmetry; and we can define the symmetry on the pentagon figure by applying the group to the set of vertices of the pentagon. For example, there's one group which describes rotational symmetries; for a pentagon, it's a cyclic group (I'll talk about what cyclic groups are later) with 5 elements; each element corresponds to one of the five possible symmetric rotations of the pentagon.

• Problem with the definition:

An action of a group G on a set S is a function f:GxS-->S such that....

meaning:
An element of the group induces a mapping of the set and not as written
"There's a mapping from elements of the group to elements of the set;"

the advice of one of my best instructors as a an undergrad was:
always write the function with its domain and range, otherwise it's meaningless.

By  ParanoidMarvin, at 11:51 PM

• paranoidmarvin:

I probably should have held off on posting until I cooked it a bit more. I was trying for an equivalent formulation that was a bit more intuitive. The standard definition is based on viewing the group through the lens of its isomorphic subgroup of the symmetric group - so that the members of the group become maps; then the group applies directly on the set, because the group itself is a function on the set. But if you're looking at a group from a different angle, the members of the group aren't a function - they're just a set of values; it can be a big leap from group members as simple values to all members of all groups being viewable as maps. I wanted to delay getting to that; looking back, I shouldn't have. I'll probably edit the post later today to change that presentation. (Don't have time now; stuff that pays the bills takes precedence.)

By  MarkCC, at 9:49 AM

• Your definition of a normal subgroup states: "all x in G : (all y in S : x * y * x^-1 = y)".

This should read: "all x in G : (all y in S : x * y * x^-1 in S)". The way you have it, it says that every element in S commutes with every element in G, which is much stronger than saying that S is normal.

It's there for a reason :-)

I know that I blew it in a couple of places, where I was trying to write things in a way that appeal to intuition. As it stands now, that post is a demonstration of the fact that it's a very bad idea to try to write intuition-based formulations of math when you're in need of sleep.

By  MarkCC, at 11:44 AM

• There is another error, but now it's a typo:

S is a function f such that:
all g,h in G : (all a in A : f(g*h,a) = f(g,f(h,a)))
all a in A : f(1,a) = a. (where 1 is the group identity element).

You change the set name from S to A in the definition.

BTW, in my opinion, a much more intuitive way of explaining this, and one which I use with my students, is to think about group actions as a machine which swallows a group element and a set element and spits out a set element, with some rules.

By  ParanoidMarvin, at 10:24 AM

• marvin:

Thanks for pointing out the typo.

I really like the "machine" metaphor for group actions. Probably my CS background, but I frequently find machine-like explanations of things to be the easiest to understand. I'd not heard of that description of a group action before.

By  MarkCC, at 10:10 AM