Find the order of an element of finite group Let $G$ be a finite group and $g,h\in G-\{1\}$ such that $g^{-1}hg=h^2$.
In addition $o(g)=5$ and $o(h)$ is an odd integer. Find $o(h)$.
I know from a previous exercise that if there exists a natural number $i$ such that $g^{-1}hg=h^i$ then for all $n\in \mathbb{N}$, $g^{-n}hg^n=h^{i^n}$.
I thought I could use this fact somehow, but so far no luck.
Please give me a hint.
 A: Square both sides
$$
g^{-1}h^2 g = h^4
$$
Now replace $h^2$
$$
g^{-1}(g^{-1} h g) g  = g^{-2} h g^2 = h^4
$$
Again square and expand
$$
g^{-3}hg^3 = g^{-2}h^2g^2 = h^8.
$$
By repeating this process we find
$$
g^{-k} h g^k = h^{2^k}
$$
for $k>0$
so in particular
$$
h = g^{-5}hg^5 = h^{32}.
$$
Hence $h^{31} = 1$. Thus $O(h)$ divides $31$. But $31$ is prime, so $O(h) = 1$ or $O(h) = 31$. Since we assumed $h \neq 1$, we find $O(h) = 31$.
Note here that the given information $O(h)$ is an odd integer was an unnecessary assumption. We could immediately see this by noting that the order of $h$ and any conjugate, e.g. $g^{-1}h g = h^2$, are the same. Thus $h,h^2$ have the same order, so $2$ is not a divisor of $O(h)$. 
A: try finishing off this line of argument...
$$
\begin{align}
h &= ehe \\&= g^{-5}hg^5 \\&= g^{-4} (g^{-1}h g) g^4 \\&= g^{-4}h^2 g^4 
\\&=g^{-3}(g^{-1}h^2g)g^3 \\&=g^{-3}h^4g^3 \\&= \dots \\&=h^?
\end{align}
$$
so $h^{?-1}=e$
A: $h=g^{-5}hg^5=g^{-4}h^2g=g^{-3}h^4g^3=g^{-2}h^8g^2=g^{-1}h^{16}g=g^{32}$.
So $g^{31}=e$. Since $31$ is prime the order is $31$.

I used $g^{-1}h^n g=(g^{-1}h g)^n$ which is straightforward, it is proven when we prove conjugate of product is product of conjugates, which is prove when we show conjugating by an element is an isomorphism.
A: By squaring both sides, you get $g^{-1}h^2 g = h^4$. This in turn means $g^{-2}hg^{2} = h^4$, by substituting for $h^2$ on the left hand side. You can keep doing this, until eventually you get $g^{-5}hg^5 = h^{32}$. Then $h = h^{32}$. So the order of $h$ (unless I've made some mistake) is either $1$ or $31$, since $31$ is prime. And since $1$ is ruled out as pointed out in a comment, that leaves just $31$ as the possible answer.
