Show that order of $a^k$ I have seen this everywhere but how do I show that :
if $a$ is an element of order $n$ in a group $G$ then the order of $a^k$ is $\frac {n}{(n,k)}$.
We know, since $a$ has order $n$, then $a^n = 1$
 A: We use the following lemma:
Lemma. If $a^m=e$ for some positive integer $m$, then $|a|$ divides $m$. 
Now since $\left(a^k\right)^{|a^k|} = e$, then $a^{k|a^k|} = e.$
So you know that $|a|$ divides $k|a^k|$, i.e. $k|a^k|$ is a multiple of $|a|$.
That is, $k|a^k|$ is the smallest multiple of $k$ and $|a|$.
So $k|a^k| = \text{lcm}(k, |a|)$.
But $\gcd(k, |a|) \, \text{lcm}(k, |a|) = k|a|$, so 
$$|a^k| = \frac{|a|}{\gcd(k, |a|)}.$$
A: The definition of the order of an element is :$k$ is the order of $x$ if and only if $k$ is the smallest integer such that $x^k=1$, so assume that we have an element $a$ with order $n$, now we want to prove that $\frac{n}{(n,k)}$ is the smallest element $d$ such that $(a^k)^d=1$, There is two things to do:


*

*First we must verify that  $(a^k)^{\frac{n}{(n,k)}}=1$: we have:
$$(a^k)^{\frac{n}{(n,k)}}=(a^n)^{\frac{k}{(n,k)}}=(1)^{\frac{k}{(n,k)}}=1$$
because $a^n=1$

*The second thing is to prove that it's the smallest element, suppose $(a^k)^d=1$ we want to prove that $d\geq\frac{n}{(n,k)}$, we have $(a^k)^d=1$ hence $a^{kd}=1$ and because $n$ is the order of $a$ then $n$ divides $kd$ which implies that $\frac{n}{(n,k)}$ divides $kd$ and because $(k,\frac{n}{(n,k)})=1$ using Euler's lemma we have $\frac{n}{(n,k)}$ divides $d$ finally $\frac{n}{(n,k)}\leq d$ which is what we want to prove.

A: Here is a different take.
Let $d=\gcd(k,n)$. Then $d = ku+nv$ for some $u,v \in \mathbb Z$.
Therefore, $a^d = a^{ku+nv} =  a^{ku} a^{nv} = (a^k)^u \in \langle a^k \rangle$ and so $\langle a^d \rangle \subseteq \langle a^k \rangle$.
On the other hand, since $k$ is a multiple of $d$, we have $\langle a^k \rangle \subseteq \langle a^d \rangle$.
Therefore, $\langle a^k \rangle = \langle a^d \rangle$ and so $ord(a^k)=ord(a^d)=\frac nd$.
