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Let $G$ be a cyclic group, and let $x \in G$ be its generator such that $|x| > 1$. Suppose $H$ is a nontrivial subgroup of $G$. Prove that if $m$ is the minimum positive integer such that $x^m \in H$, then $x^m$ generates $H$.

My try: Suppose $|x| = n > 1 $. We want to show that $|x^m| = |H|$. Suppose to the contrary that $|x^m| < |H|$. Since $x^n = e \implies x^{mn} = e \implies (x^m)^n = e $. So $n $ divides $m$. By division algorithm, we can take $q,r$ such that $|H| = mq + r $ where $0 \leq r < m$. Therefore

$$ x^r = x^{|H| -mq}= x^{|H|}\implies r = |H|$$

But, can we conclude then that $x^m$ is not in $H$ so that we get a contradiction?

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Is $G$ a finite cyclic group? Otherwise you can't start with $|x|=n$. – Christoph Nov 2 '13 at 10:17
The title is the only thing about the question which shows in most contexts, so it's worth having a title which gives a clue as to what the question is about. – Peter Taylor Nov 2 '13 at 10:20
up vote 3 down vote accepted

Because $x$ generates $G$, all elements of $H$ are powers of $x$. Let $h\in H$. Then $h=x^i$ for some $i$. Writing $i=mq+r$ with $0\le r < m$ we get $r=0$ by minimality of $m$. Hence all elements of $H$ are powers of $x^m$ and so $x^m$ generates $H$.

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