Injective homomorphism from finite abelian group to symmetric group

Question

This is the exercise I'm looking at:

$i)$ Is there an injective homomorphism from $\mathbb Z/10 \rightarrow S_7$?

$ii)$ Is there an injective homomorphism from $\mathbb Z/8 \rightarrow S_7$?

My problem here is that this exercise is called "cycle decomposition", but I don't know how this exercise has anything to do with it. I'm rather confused here. My first thought (before I saw the title) was to use Cayley's theorem somehow. Maybe that's a part of the way to proof it? Can someone show me the way for one of those above? Once I understand, I should be able to do the other by myself.

Thanks in advance.

Answer 1

Hint:

If there's such a map, it has to send $1$ in $\Bbb Z/10\Bbb Z $ to an element of order $10$ in $S_7$. And $2$ and $5$ therefore have to go to elements of order $2$ and $5$ respectively, and those elements must commute.

Can you think of an element $u$ of order $5$? If you're thinking of $S_7$ as permutations of $7$ items, does $u$ leave anything fixed? Can you think of an element $v$ of order $2$ that acts only on those?

Answer 2

The image of an injective homomorphism $\mathbb{Z}/n\mathbb{Z}\to S_k$ is a cyclic subgroup of $S_k$ of order $n$. Conversely, if $H$ is a cyclic subgroup of $S_k$ of order $n$, there is an injective homomorphism $\mathbb{Z}/n\mathbb{Z}\to S_k$.

So the question is: does there exist an element in $S_7$ of order $10$? And one of order $8$? How do you compute the order of a permutation in terms of the lengths of its disjoint cycles?