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I'm having doubts about an exercise, since I reach a trivial conclusion. I hope it's not an issue to see if I'm screwing up.

Let $G$ be a group, and $A$ a normal abelian subgroup. Show $G/A$ operates on $A$ by conjugation and in this way get a homomorphism of $G/A$ into $\text{Aut}\ (A)$.

So the action is $gA\bullet a=gAa(gA)^{-1}=gAag^{-1}A=gAaAg^{-1}=gAg^{-1}=A$. Also $(gAhA)\bullet a=gA\bullet(hA\bullet a)$ and $A\bullet a=A$. So is this just a trivial homomorphism which maps everything to $A$?

I think I'm applying things wrong since I thought it's supposed to be $A\bullet a=a$ and the action should return elements of $A$, not $G/A$. How to interpret correctly?

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Your description of the action is not correct. Try showing that (1) if $N$ is any normal subgroup of $G$, then $G$ acts on $N$ via conjugation; (2) if $N = A$ is abelian, then the conjugation action on $A$ is trivial when restricted to $A$, and hence that this action factors through $G/A$. – Matt E Sep 7 '11 at 19:16
Thanks, I suspected as much. – kreashawn Sep 7 '11 at 19:32
@kreashawn: This is a very good question. Everything but the first equality is actually correct. You are quite correct that the action is supposed to take elements of A to elements of A. The trivial action takes a to a, not a to A. It would not be an action if it took elements of A to elements of G/A. – Jack Schmidt Sep 7 '11 at 21:46
up vote 9 down vote accepted

Remember that any action of a group $G$ on a set $X$ induces a homomorphism $G\to S_X$, the group of permutations on $X$; and conversely, that any group homomorphism $G\to S_X$ induces an action of $G$ on $X$. Given an action $G\times X\to X$, the homomorphism $\varphi\colon G\to S_X$ maps $g$ to $\varphi(g)$, the function such that $\varphi(g)(x) = g\cdot x$. Conversely, given a homorphism $\psi\colon G\to S_X$, $\psi$ defines an action by $g\cdot x = \psi(g)(a)$.

With that preliminary, let let $G$ act on $A$ by conjugation in the usual way: $g\cdot a= gag^{-1}\in A$ (it's in $A$ because $A$ is normal). This is an action of $G$ on $A$.

This induces a homomorphism $G\to S_A$, where we consider $A$ as a set.

However, because $A$ is abelian, for every $a\in A$ and every $b\in A$ we have $b\cdot a = a$. That is, the action of $A$ on itself is the identity permutation, so the homomorphism $G\to S_A$ factors through $G/A$. This gives a homomorphism $G/A \to S_A$. And of course, any homomorphism from a group $H$ to a permutation group $S_X$ induces an action of $H$ on $X$, so $G/A$ acts on $A$.

The action of $G/A$ on $A$ is as follows: given $gA\in G/A$ and $a\in A$, $gA\cdot a = gag^{-1}$. This is well defined, because if $gA=hA$, then $h^{-1}g\in A$, so $(h^{-1}g)a(g^{-1}h) = a$, hence $gag^{-1}=hah^{-1}$.

The action is not trivial unless $A$ is central.

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As you say, what you've written doesn't make sense: if $G/A$ acts on $A$ then it maps elements of $A$ to elements of $A$. But you started with $a \in A$, acted by $gA \in G/A$, and ended up with $A$. Something is wrong here :). The right thing to do is to define the action of $gA \in G/A$ on $a\in A$ by $$ gA \cdot a := ga g^{-1} $$

There's an issue of well-definedness here: the action seems to depend on a choice of coset reps $g$. You need to check a different choice of coset rep $g'$ doesn't make a difference to the action: i.e. $ga g^{-1} = g' a g'^{-1}$.

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$\alpha$′(gA)(a)=a, then g belong to Z is not correct, maybe any element which can commute with a.

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What doy ou mean by $\alpha^\prime$? – draks ... Aug 2 '13 at 9:25

Let $\alpha$: G-->Aut(A) is a homomorphism, then if $\alpha$(g)(a)=ga$g^{-1}$=a, we can get that g belong to A, so ker$\alpha$=A, from this point we obtain $\alpha^{'}$ which is induced by $\alpha$:G/A-->Aut(A), which is a injective inner aut.

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There is no reason why only elements of $A$ acts trivially on $A$. It is true however, that the kernel contains $A$, which is what you need for the action to factor through $G/A$, but the map from $G/A$ to $\rm{Aut}(A)$ will not in general be injective. – Tobias Kildetoft Aug 1 '13 at 9:02

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