For what $n$ is there an injective homomorphism from $\mathbb{Z_n} \to S_7$

So I came across this problem:

For what $n$ is there an injective homomorphism from $\mathbb{Z}_n\to S_7$?

I have the solution but the solution doesn't make much sense to me. It said that $n=1,2,3,4,5,6,7,10,12$. Also I'm wondering why $n$ can't be $8$?

A homomorphism $\phi$ is just a mapping such that for all $a,b\in \mathbb{Z}_n$ $\phi(ab)=\phi(a)\phi(b)$

Since we can always map the elements of $\mathbb{Z}_8$ injectively to $S_7$ as the order difference is quite large I would assume that the property $\phi(ab)=\phi(a)\phi(b)$ is the property that doesn't hold for $\mathbb{Z}_8$.

But I am not exactly sure how to explicitly show this, please help!!

Solution 1:

The question is equivalent to asking whether we can find an element $a \in S_7$ with order $n$. (If we can, then we map $1\in \mathbb Z_n$ to $a$, which uniquely determines the mapping; can you see why it must be injective? Conversely, if we have an injective mapping $\phi$, $\phi(1)$ must have order $n$.)

So is there any element of $S_7$ of order $8$?

The elements of $S_7$ are permutations. One way to write a permutation is to decompose it into disjoint cycles. The best way to explain this is to illustrate by example: one element of $S_7$ is written $(1436)(27)(5)$, which is the permutation that sends $1$ to $4$, $4$ to $3$, $3$ to $6$, and $6$ back to $1$; $2$ to $7$ and $7$ to $2$; and $5$ to itself. That's a $4$-cycle, a $2$-cycle, and a $1$-cycle.

There's a way to find the order of an element based on the lengths of the cycles in its cycle decomposition. I'm not going to give away the punchline yet. But play with some examples of these cycle decompositions and see if you can find their orders.