Prove for two groups, one less than the other, the smaller is a cyclic subgroup of larger.

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.

• How many elements do you need to generate $H$ given the if condition? – Justin Benfield Mar 13 '16 at 18:20

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