Prove that $\pi$ is an action.

Question

Question. Let $G$ be abelian; and define an action of $\pi$ of $G$ on $G$ by $$\pi_g(x)=g^2x.$$

Prove that $\pi$ really is an action.

Attempt. Let $g$ be the identity element; then $\pi_1(x)=x$ so the identity condition for an action holds.

Now, for $f,g\in G$ we have $$\pi_{fg}(x)=(fg)^2x=fgfgx=ffggx=f^2g^2x=\pi_f(x)\pi_g(x),$$ so $\pi$ really is an action as desired.

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The only problem I have with this is $$\pi_f(x)=f^2x,\quad \pi_g(x)=g^2x,$$ so is it true that $\pi_f(x)\pi_g(x)=f^2xg^2x$? Then in that case, I guess my attempt is wrong since I don't have that extra $x$; so where have I gone wrong?

Thanks in advance.

Answer 1

You have (almost) correctly proved $\pi$ is an action. Where you write $$ f^2g^2x=\pi_f(x)\pi_g(x) $$ you should write $$ f^2g^2x=\pi_f(\pi_g(x)) . $$ What your second observation about the "extra $x$" shows is that $\pi$ is not a group automorphism. But there's no reason why it should be.

Answer 2

$f^2g^2x\ne \pi_f(x)\pi_g(x)=f^2xg^2x=f^2g^2x^2$. And that is not a bad thing, because you do not need to prove that for all $x$, $\pi_{fg}(x)=\pi_f(x)\pi_g(x)$, but rather that $\pi_{fg}=\pi_f\circ \pi_g$: id est for all $x$, $\pi_{fg}(x)=\pi_f(\pi_g(x))$.