Let $\phi_a$ be the automorphism of $G$ given by $\phi_a(x) = axa^{-1}$. Show that $|\phi_a|$ divides $|a|$.

Question

Assumption: Let $a$ belong to a group $G$ and let $|a|$ be finite.
I proceeded as follows:
Let $|a|=m.$We have $\phi_a (x)=axa^{-1}$ and so $(\phi_a(x))^2=(axa^{-1})(axa^{-1})=ax(a^{-1}a)xa^{-1}=ax^2a^{-1}\implies (\phi_a(x))^m=ax^ma^{-1}$.
From here, how do I show that $(\phi_a(x))^m=x$? Since, $e(x)=x$, we'll have $(\phi_a)^m=e$ and the result will follow. Note that $e$ is the identity.

Answer 1

Hint: $$\begin{align} \phi_a^2(x)&=\phi_a(\phi_a(x))\\ &=\phi_a(axa^{-1})\\ &=a(axa^{-1})a^{-1}\\ &=a^2x(a^{-1})^2. \end{align}$$ Let $H_a$ the group generated by $a$ there exists a morphism $f:H\rightarrow {\rm Aut}(G)$ defined by $f(x)=\phi_x$, let $K$ be the kernel of $f$, $|{\rm Im}(f)||{\rm Ker}(f)|=|H|$.

Answer 2

Hint: More generally, if $\phi: G \to H$ is a homomorphism of groups and $a \in G$ has finite order, then the order of $\phi(a)$ is finite and divides the order of $a$. Apply this to $\phi: G \to Aut(G)$ given by $\phi(a)=\phi_a$.