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?