Why $\mathbb{Z_{12}}$ is subgroup of $S_{7}$?

Question

In one the solutions of my Abstract algebra question, It was written,

$\mathbb{Z_{12}}$ is not a subgroup of $S_{6}$.

$\mathbb{Z_{12}}$ is a subgroup of $S_{7}$.

$\mathbb{Z_{12}}$ is a subgroup of $S_{9}$ and

$\mathbb{Z_{12}}$ is a subgroup of $S_{11}$.

But my question is, since this groups $S_{6}$,$S_{7}$, $S_{9}$ and $S_{11}$ are non abelian group, how can we be sure whether it has $\Bbb Z_{6}$ is its subgroup or not?

Thanks in advance..

Answer 1

HINT: What is the order of the permutation $(1234)(567)$?

Answer 2

It helps to realize that $\Bbb Z_{12} \cong \Bbb Z_3 \times \Bbb Z_4$. If you look at it that way, it's easy to see how to inject $\Bbb Z_{12}$ into $S_7$.