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Suppose that $H$ and $G$ are groups and that $H \le G$. Prove that if $H \cong \mathbb Z$ or $H \cong \mathbb Z_n$, then $H=\langle g \rangle$ for some $g \in G$

I'm not entirely sure where to go on this problem. I guess if $H= \mathbb Z$, then if for $g=1$, then $H= \langle g \rangle = \langle 1 \rangle$. However, I don't know how to make a general argument.

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  • $\begingroup$ How many elements do you need to generate $H$ given the if condition? $\endgroup$ – Justin Benfield Mar 13 '16 at 18:20
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$\Bbb Z$ and $\Bbb Z_n$ are cyclic. If $H$ is $\cong$ to one of them, then it is cyclic. Consequently there is some $g \in H$ such that $H = \langle g \rangle$. But $H \subset G$, so $g \in G$.

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  • $\begingroup$ Is $\mathbb Z$ considered cyclic because it is just one large set unlike $\mathbb Z_n$? $\endgroup$ – AndroidFish Mar 13 '16 at 18:28
  • $\begingroup$ @AndroidFish $\mathbb Z$ is cyclic because it is equal to $\langle 1 \rangle$. $\endgroup$ – user258700 Mar 13 '16 at 18:28

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