How to show that $\mathbb{Z}_{12} $ is isomorphic to a subgroup of $S_7$?

Question

How to show that $\mathbb{Z}_{12}$ is isomorphic to a subgroup of $S_7$?

My attempt: Using Cayley's theorem one can conclude $\mathbb{Z}_{12}$ is isomorphic to a subgroup of $S_{12}$.

Or, if I use Generalised Cayley's theorem I can show that there is a homomorphism from $\mathbb{Z}_{12}\rightarrow S_{\mathbb{Z}_{12}/H}$ where $H$ is a subgroup of order $3,2^2$, therefore we have group homomorphism from $\mathbb{Z}_{12}$ to $S_3$ or $S_4$. But these maps have non-trivial kernel namely $H$ itself.

Therefore, I have not been able to conclude the required statement.

Any help is appreciated.

Answer 1

Hint: look at $(1 \ 2 \ 3)(4 \ 5 \ 6 \ 7)$. What is its order?

Answer 2

In this case, as Nicky Hekster pointed out in the top answer, you know the general structure of elements in $S_7$ so finding one of order $12$ is pretty easy.

I want to add that a more general procedure to look for a copy of $\mathbb{Z}_{12}$ in a group $G$ is to consider that $\mathbb{Z}_{12} = \mathbb{Z}_4 \times \mathbb{Z}_3$. So you can look for an element of order $3$ (computing the Sylow $3$-subgroup) and then seeing if the centraliser of this element has an element of order $4$ (or, the dual procedure, looking for a copy of $\mathbb{Z}_4$ in the Sylow $2$-subgroup of $G$ and then computing the order of its centraliser, if $3$ divides it you are done).

Of course, the difficulty in this lies in the fact that computing Sylow subgroups or centralisers might not be easy, but in some cases it is. (First example I can think of, if $G=S_4 \times S_4$ you can take $((1234), 1)$ as the element of order $4$ and it is immediate that $(1,(123))$ is contained in its centraliser).