Problem source

Let $G$ be a finite group with identity $e.$ For each element $g\in G,$
the order of $g$, $o(g),$ is defined to be the smallest positive
integer $n$ for which $g^{n}=e.$
\begin{questionparts}
\item Show that, if $o(g)=n$ and $g^{N}=e,$ then $n$ divides $N.$
\item Let $g$ and $h$ be elements of $G$. Prove that, for any integer
$m,$ 
\[
gh^{m}g^{-1}=(ghg^{-1})^{m}.
\]
\item Let $g$ and $h$ be elements of $G$, such that $g^{5}=e,h\neq e$
and $ghg^{-1}=h^{2}.$ Prove that $g^{2}hg^{-2}=h^{4}$ and find $o(h).$ 
\end{questionparts}

Solution source

\begin{questionparts}
\item Show that, if $o(g)=n$ and $g^{N}=e,$ then $n$ divides $N.$

Using the division algorithm, write $N = qn + r$ where $0 \leq r < n$ to divide $N$ by $n$. Then we have $e = g^N = g^{qn + r} = g^{qn}g^r = (g^{n})^qg^r = e^qg^r = g^r$ therefore $r$ is a number smaller than $n$ such that $g^r = e$. Therefore either $r = 0$ or $o(g) = r$, but by definition $o(g) = n$ therefore $r = 0$ and $n \mid N$.

\item Let $g$ and $h$ be elements of $G$. Prove that, for any integer
$m,$ 
\[
gh^{m}g^{-1}=(ghg^{-1})^{m}.
\]

$(ghg^{-1})^m  = \underbrace{(ghg^{-1})(ghg^{-1})\cdots(ghg^{-1})}_{m \text{ times}} = gh(g^{-1}g)h(g^{-1}g)\cdots (g^{-1}g)hg^{-1} = gh^mg^{-1}$

\item Let $g$ and $h$ be elements of $G$, such that $g^{5}=e,h\neq e$
and $ghg^{-1}=h^{2}.$ Prove that $g^{2}hg^{-2}=h^{4}$ and find $o(h).$ 

$g^2hg^{-2} = g(ghg^{-1})g^{-1} = gh^2g^{-1} = (ghg^{-1})^2 = (h^2)^2 = h^4$.

$h = g^{5}hg^{-5} = g^4ghg^{-1}g^{-4} = g^4h^2g^{-4} = g^3(ghg^{-1})^2g^{-3} = g^3h^4g^{-3} = h^32$

Therefore $e = h^{31}$. Therfore $o(h) \mid 31 \Rightarrow \boxed{o(h) = 31}$ since $31$ is prime and $o(h) \neq 1$