Here is the problem.
Let $|G|=n,$ and let $m$ be an integer with $\operatorname{gcd}(m, n)=1 .$ Define $\sigma: G \rightarrow G$ by $\sigma(g)=g^{m}$ for all $g \in G .$ If $G$ is abelian, show that $\sigma$ is an automorphism of $G$.
I proved that $\sigma$ is bijection. For the rest, I tried as follows:
Let $G$ be an abelian group with $|G|=n$. Let $m$ be an integer with $\operatorname{gcd}(m, n)=1 .$ Define $\sigma: G \rightarrow G$ by $\sigma(g)=g^{m}$ for all $g \in G .$ We prove that $\sigma$ is an automorphism of $G$. By Corollary of Lagrange's Theorem, we observe that $g^{n}=1$ for all $g \in G$.
Now we show that $\sigma$ is an isomorphism. Let $g, h \in G$. If $m \geq 0$, we have
$$
\sigma(gh)=(gh)^{m} =g^{m} h^{m}= \sigma(g)\sigma(h).
$$
If $m<0$, there is $c \in \mathbb{N}$ such that $m=-c$. Then we see that
$$\begin{align}
\sigma(gh)&=(gh)^{m} \\
&=(gh)^{-c}\\
& = ((gh)^{-1})^{c}\\
&=(h^{-1}g^{-1})^c\\
&=h^{-c}g^{-c}\\
&=h^{m}g^{m}\\
&=g^{m}h^{m}\\
&= \sigma(g)\sigma(h).
\end{align}$$
Hereby, $\sigma$ is an isomorphism.
Here is my question:
1) Is the proof right?
2) Did I use correctly the condition that $G$ is abelain?
Thanks!!
Best Answer
Yes, your proof is fine. So is your use of the hypothesis that $G$ is abelian.
You could be more specific about which corollary to Lagrange's Theorem you're using, although I think it's implicit.