Let $a$ be an element of order $n$ in a group $G$. If $a^m$ has order $n$, then $m$ and $n$ are relatively prime.

Let $a$ be an element of order $n$ in a group $G$.

If $a^m$ has order $n$, then $m$ and $n$ are relatively prime.

Assume $a^m$ has order $n$ and, $m$ and $n$ are not relatively prime. Then $m$ and $n$ have a common factor, say $q$. So, $m=m'q$ and $n=n'q$. So,

$$(a^m)^{n'}=(a^m)^\frac nq= (a^{mn})^\frac 1q= e^\frac 1q=e$$

So, $(a^m)^\frac nq=(a^m)^n=e$.

Since $a^m$ has order $n$: $e \neq (a^m)^\frac nq = (a^m)^n$. Contradiction.

Solution 1:

Your idea is good, but you can't do $g^{\frac{1}{k}}$ in groups.

Let $d=\gcd(m,n)$ and write $m=m'd$, $n=n'd$. Then $$ (a^m)^{n'}=(a^{m'd})^{n'}=a^{n'dm'}=(a^{n'd})^{m'}=(a^n)^{m'}=e^{m'}=e $$ Since $n'\le n$ and $a^m$ is assumed to have order $n$, we conclude that $n'=n$, so $d=1$.