Is this proof that group elements in the same conjugacy class have the same order incomplete? My question is similar to the question asked here: If two elements belong to the same conjugacy class then they have the same order, but none of the answers there seem to answer what is confusing me.
My lecture notes give the following proof:
Let $a$ and $b$ be two elements in group $G$ that are in the same conjugacy class. We also let $n$ be the smallest possible integer s.t. $a^n=e$, where $e$ is the identity element in $G$. An arbitrary conjugate $b$ of $a$ is $b=gag^{-1}$, for some element $g$ in $G$. Then,
$$b^n=(gag^{-1})(gag^{-1})(gag^{-1})...(gag^{-1})=ga^ng^{-1}=geg^{-1}=e$$
The proof then concludes that this is sufficient to show that $b$ is also of the order $n$. But wouldn't it also be necessary to prove that there is no $k<n$ s.t. $b^k=e$, i.e. to me this proof only shows that there $b$ is at most the same order as $a$, but it does not prove that it can't be of a smaller order. Why is this not an issue then?

Slightly different approach: for any $g \in G$ let the map $\phi_g : G \rightarrow G$ be defined by $\phi_g(x)=gxg^{-1}$. It is easy to check that this is an automorphism of $G$. Hence in your case, $b=\phi_g(a)$ has the same order as $a$.
