I'm trying to show that for $\phi \in Aut(G)$ and $g \in G$, if $\exists n \gt 0$ such that $x^n = 1_G$ then $\phi$ preserves the order.

In other words, if $\exists n$ such that $x^n = 1_G$ then $(\phi(x))^n = 1_G$.

It seems very obvious to me and I'm thinking I might be missing something. Since $\phi \in Aut(G)$ it is an isomorphism from G to G and as such:

$$ (\phi(x))^n = \phi(x^n) = \phi(1_G) $$

Since $\phi$ is an isomorphism, $Ker(\phi) = \{1_G\}$ and so we get $1_G$ above.

Does that seem right? Thanks!


1 Answer 1


If $x$ has order $n$ then $\phi(x)^n=\phi(x^n)=\phi(1)=1$ so the order of $\phi(x)$ divides $n$. Since $\phi$ is an automorphism, we have an inverse, so letting $y=\phi(x)$ have order $m$ and applying $\phi^{-1}$ as above we get $n|m$. But the first statement implies $m|n$ and hence we have $m=n$, i.e. they have the same order.


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