Suppose $G$ is a group and $a\in G$ with $|a| = m$. Prove that $\langle a^k \rangle=\langle a \rangle \iff \gcd(k, m) = 1$. Suppose $G$ is a group and $a\in G$ with $|a| = m$. Prove that $\langle a^k \rangle=\langle
a \rangle \iff \gcd(k, m) = 1$.
 A: $\langle a^k\rangle = \langle a\rangle \iff a \in \langle a^k\rangle\iff \exists n\!: a^{kn}\!=a \iff\exists n\!: a^{kn-1}=1\overset{{\rm ord}\,a\, =\, m}\iff\exists n\!:\ m\mid kn\!-\!1$ $\!\iff\!$ $\exists j,n\!:\ jm+kn = 1\iff \gcd(m,k)=1\,$ by Bezout
A: If $\langle a^k\rangle=\langle a\rangle$, then there exists $n$ such that $a^{kn}=a$, so that $kn\equiv 1\pmod{m}$. Thus $k$ is invertible mod $m$, and therefore $\gcd(k,m)=1$.
Suppose conversely that $\gcd(k,m)=1$. We know that for all $n$ we have $a^{kn}\in\langle a\rangle$. Thus $\langle a^k\rangle\subseteq \langle a\rangle$. To show the opposite containment it will suffice to show that there is some $n$ such that $a^{kn}=a$. This will happen if $kn\equiv 1\pmod{m}$, and since $\gcd(k,m)=1$ there exists an $n$ making this happen. Thus the opposite containment holds.
A: For another method suppose $ \langle a^k \rangle = \langle a \rangle $ and  $ d = \gcd(k, m) \gt 1 $. 
Then there is $t, s \in \Bbb Z$ such that  $dt =  m$ and $ds = k$ with $t \lt m$. 
Then, $ (a^k)^t = (a^{ds})^t = (a^{dt})^s = (a^m)^s = e^s = e  $. But then $ \langle a^k \rangle $ would have less then $t$ elements, hence less than $m$ elements whence $ \langle a^k \rangle \neq \langle a \rangle $ leading to a contradiction. 
Hence $\gcd (k,m) = 1$. 
For the other direction, suppose $k, m $ are relatively prime integers. Then there exists integers $x, y $ such that $ mx + ky  = 1$. 
Then for any integer $n$ we have that $ n = nmx + nky  $. Hence, $$ a^n = (a^m)^{nx} \cdot (a^k)^{ny} = (a^k)^{ny}$$ which establishes that  $ \langle a^k \rangle \supseteq \langle a \rangle $. The other inclusion is obvious. 
$\mathscr{Q.E.D.}$
