What are the possible values for n if there is an injective homomorphism from $Z_{n}$ to $S_{7}$

Question

I am guessing that $Z_{7}$ is the only $Z_{n}$ who can have injective homomorphism to $S_{7}$ by Cayley theorem. My mapping is for any $x\in Z$, $f(x) = \begin{bmatrix}x+0&x+1 &x+ 2 &x +3 &x+4 &x+5 &x+6\\ 0&1 &2 &3 &4 &5 &6\end{bmatrix} $. This mapping is isomorphism so it should be injective homomorphism too.

However I have no idea how to argue that this is the only $Z$ group injective homomorphism.

Or is there any other $Z_{n}$ which is injective homomorphism to $S_{7}$?

Answer 1

Hint: There is an injective homomorphism from $\mathbb Z_n$ into $S_7$ if and only if there is an element $g\in S_7$ whose order is $n$; just consider the map from $\mathbb Z_n$ into $S_n$ defined by $k\mapsto g^k$. So, what are the possible orders of element of $S_7$?

Answer 2

Hint: Since $\mathbb{Z}/n\mathbb{Z}$ is cyclic and you want to define an isomorphism onto the image, you need to figure out which cyclic subgroups $S_7$ has.

Edit: No, not only $\mathbb{Z}/7\mathbb{Z}$ can be embedded into $S_7$. There will be $9$ different $n$ that work.