A little concept of group actions.

I know that given a group G and a set S, a group action of G on the S is a function F:S ---> S satisfying some properties.

But what really confuses me arises from a question:

Suppose G is a group and A is an abelian normal subgroup of G, how does G/A operates on A by conjugation?

First, "operates" means "acts"?

If so, then given an element gA of G/A, is the action like... gA$\displaystyle .$a=(gA)a(gA)^{-1}=(gA)a(g^{-1}A)=gAaAg^{-1}=gAg^{-1}=A ?

See, this is where it confuses me.

Group actions by conjugation should take its element gA instead of the basic element ga, but what it follows is that I got a set A as the result of the action instead of an element of A.

I did it by its basic definition, but ended up with a set? Can anyone tell me what's wrong with my equation?

Re: A little concept of group actions.

Hi Rita,

I think your work was heading in the right direction. I will try to formalize what you were saying below.

Let $\displaystyle \cdot$ be the group multiplication. G/A acts on A in the following way: Let $\displaystyle gA\in G/A.$ Then the action $\displaystyle \star: G/A \times A\rightarrow A$ is __defined__ by $\displaystyle gA\star a=g\cdot a\cdot g^{-1}.$ Note that the definition of $\displaystyle \star$ is legitimate, because $\displaystyle A$ is normal in $\displaystyle G$ so $\displaystyle g\cdot a\cdot g^{-1}\in A.$ The tough work is showing that $\displaystyle \star$ is well-defined. That $\displaystyle \star$ is well-defined follows from the fact that $\displaystyle A$ is abelian.

Does this help clear things up? Let me know if anything is unclear. Good luck!

Re: A little concept of group actions.

here is what we need to show:

that if gA = g'A, we get the same conjugate of a if we conjugate by g or g'.

recall that two cosets gA and g'A are equal if and only if g'^{-1}g is in A.

so g'^{-1}g = a', for some a' in A. therefore:

g = g'a'. now pick an arbitrary element a in A. then:

gag^{-1} = (g'a')a(g'a')^{-1} = g'(a'aa'^{-1})g'^{-1}.

but A is abelian, so a'aa'^{-1} = aa'a'^{-1} = ae = a. hence:

gag^{-1} = g'(a'aa'^{-1})g'^{-1} = g'ag'^{-1}, as desired.

EDIT:

here, our set S is A = {e,a_{1},a_{2},.....} (the subscripts aren't "really fair" A might be uncountable, so we might need a larger index set than the natural numbers).

given an element of a group G/A we need to come up with a bijective mapping A-->A "induced" by an element of G/A.

the mapping in this case is a--->gag^{-1}, so (gA).a = gag^{-1}.

we need A normal in G to be sure that gag^{-1} is in A (or else we don't have a mapping from A to A). we know this mapping is bijective, because it's an element of Inn(G), and inner automorphisms are bijective (even when restricted to a subset A of G).

we need A to be abelian, because that is what guarantees that it doesn't matter which "g" we pick, any g in the coset gA will gives the same conjugate of A.

let's pick a non-abelian group G, with a normal abelian subgroup A, and see how one of these things might work:

we'll pick G = S_{3}, and A = {e,(1 2 3), (1 3 2)} = A_{3}. the elements of G/A are {A, (1 2)A} (G/A has order 2).

so we are going to define (gA).a = gag^{-1}.

for example, A.a = a for any element a of {e, (1 2 3), (1 3 2)} (because A = eA). it's pretty clear that since A is abelian, conjugating an element of A with another element of A just gives back our original element.

the interesting part is what happens when we let (1 2)A act on A:

we can conjugate by any element of (1 2)A = {(1 2), (2 3), (1 3)}. let's conjugate by (1 2) and see what happens:

(1 2)e(1 2)^{-1} = (1 2)e(1 2) = e.

(1 2)(1 2 3)(1 2) = (1 2)(1 3) = (1 3 2)

(1 2)(1 3 2)(1 2) = (1 2)(2 3) = (1 2 3)

note this is the same map we get if we conjugate by (2 3) instead:

(2 3)e(2 3) = e

(2 3)(1 2 3)(2 3) = (2 3)(1 2) = (1 3 2)

(2 3)(1 3 2)(2 3) = (2 3)(1 3) = (1 2 3)

if we write A = {e,a,a^{2}} (where a = (1 2 3)) then:

A (the coset) acts on A (the set) as the identity map:

e-->e

a-->a

a^{2}-->a^{2}

(1 2)A (the coset) acts on A (the set) as the mapping:

e-->e

a-->a^{2}

a^{2}-->a

in other words, any element of A_{3} induces the identity map of A_{3}, and any 2-cycle induces the inversion map on A_{3}. (which switches a and a^{-1}).

be careful! we can't ALWAYS let a quotient group act on a subgroup. if we have a quotient group G/N, we can get a mapping from N-->N by letting G act on N by conjugation (since N is normal), but there is no guarantee that we can get an action of G/N on N (interestingly enough we CAN get an action of a subgroup H of G on N by conjugation, though), unless N is abelian, too.

Re: A little concept of group actions.

Oh! So the key point is how the conjugation forms. It maks sense a lot, thank you soooo much. :)