Let $(G,*)$ be a group and $a in G$.
Suppose that $|a|=n$ and $n=mk$ for some positive integers $m$ and $k$.
What is $|a^k|$?
attempt:
Let $a in G$ such that $|a|=n$. Then,
begin{equation*}
a^n = a^{mk} = (a^k)^m = e_G
end{equation*}
where $e_G$ is the identity element of $G$. Thus, the order of $a^k$ is $m$.
In particularly, by theorem:
Let $(G,*)$ be a group, $a in G$, and
$|a|=n$. Then, for all $k in Bbb N$,
$|a^k| = frac{n}{gcd(k,n)}$,
we have, $|a^k| = frac{n}{gcd(k,mk)} = frac{n}{k} = m$.
Does it true? On the other hand, in the answer key says that the answer is
$frac{n}{m}$. How to get this?
