Let $G$ denote a group and let $a,x \in G$. Eventually, I want to show that the orders $|xax^{-1}|=|a|$.
I was thinking of splitting the proof into two cases: One where $|a|$ is finite, and the other where $|a|$ is infinite.
However, the book only considers the finite case. Is this the only case that must be considered?
Hint: For any $x\in G$, the function $\varphi_x: G \to G$ given by $\varphi_x(a) = xax^{-1}$ is an automorphism. You can check that it's a group homomorpshism on your own. Also, $\ker{\varphi_x}=\{e\}$ because $\varphi_x(a)=xax^{-1}=e \implies a=x^{-1}x=e$ which means that it's injective and it is surjective because for any $y \in G$, $\varphi_x(x^{-1}yx)=x\big(x^{-1}yx\big)x^{-1}=y$.
Now use the fact that a group isomorphism preserves the order of elements. In other words, $|xax^{-1}|=|\varphi_x(a)|=|a|$. To get started, note that if you have a homomorphism $\psi: G_1 \to G_2$, then $\big(\psi(a)\big)^{o(a)}=\psi(a^{o(a)})=\psi(e_{G_1})=e_{G_2}$ which implies that $o\big(\psi(a)\big) \mid o(a)$. Now use injectivity to show that $o(a) \mid o\big(\psi(a)\big)$ too.
The advantage of this method is that you don't need to worry whether $G$ is a finite or an infinite group. Also, if $o(a)=\infty$, then $\infty \mid o\big(\varphi_x(a)\big)$ which means that the order of $\varphi_x(a)$ must be infinite too because $n \mid m \implies |n| \leq |m|$.