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Let $G$ be a finite cyclic group generated by $x$. $(|G|=k)$

Let $n,m\in\mathbb{Z}$ such that $\gcd(n,k)=\gcd(m,k)$.

Then, $\langle x^n\rangle=\langle x^m\rangle$.

I can prove the converse, but i don't know how to prove this one.

I can show that $|\langle x^n\rangle|=|\langle x^m\rangle|$ but how does this imply that these two sets are equal?

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  • $\begingroup$ You can try showing that $x^m\in\langle x^n\rangle$ and vice versa. $\endgroup$ – Nishant Apr 15 '14 at 1:19
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Let $d=\gcd(m,k)=\gcd(n,k)$. We show that $<x^m>=<x^d>$ and $<x^n>=<x^d>$.

Only one of these needs to be shown. It is obvious that every element of $<x^m>$ is a power of $x^d$. So we need only show that $x^d$ is a power of $x^m$.

Since $\gcd(m,k)=d$, there exist integers $s$ and $t$ such that $ms+kt=d$. It follows that $$x^d=x^{ms+kt}=(x^m)^s (x^k)^t=(x^m)^s,$$ and we are finished.

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  • $\begingroup$ Common gcd of what? Gcd of $m,k$?? $\endgroup$ – John. p Apr 15 '14 at 1:30
  • $\begingroup$ I will edit to clarify. $\endgroup$ – André Nicolas Apr 15 '14 at 1:31
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Hint $\ $ The set $S$ of integers $j$ such that $\,x^j \in \langle x^n\rangle$ is closed under subtraction, so closed under gcd. Thus $\,n,k\in S\,\Rightarrow\,(n,k)\in S.\,$ Thus $\, \langle x^n\rangle \supseteq \langle x^{(n,k)}\rangle,\,$ and reversely by $\,(n,k)\mid n.\,$ Similarly $\,\langle x^m\rangle= \langle x^{(m,k)}\rangle\,$ thus $\langle x^m\rangle= \langle x^{\color{#c00}{(m,k)}}\rangle=\langle x^{\color{#c00}{(n,k)}}\rangle =\langle x^n\rangle$ by $\,\color{#c00}{(m,k)=(n,k)}\ \ $ QED

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