### 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

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.)

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

A group S is a subgroup of G iff:

There is a particular kind of subgroup that is interesting, called a

Another way of writing that is:

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

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

A

A

A

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

If you have a group G and a set A, then a

*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)

all x in G : x * S * x^-1 is a subset of S

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).

- There's a mapping from elements of the group to functions over the set;
- Performing the group operation on the elements of set according to the mapping applies a symmetric transformation to the set.

*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.
## 6 Comments:

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

allmembers 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.

By Adam, at 11:30 AM

adam:

Notice the header that says "This article is being rewritten, don't waste your time reading it now"?

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

verybad 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

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