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

Question

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!!

Answer 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.

Answer 2

For n = 2,3,4,5,6,7,10,12 you can get an injective group homomorphism from Zn into S7. Let f : Zn --> S7 be an injective group homomorphism. Then |Ker f| = 1 as f is injective and Zn is isomorphic to a subgroup ( say H ) of S7. Now by isomorphism Th Zn/ Ker f ~ H and H must be cyclic as Zn is cyclic. Now in S7 there are elements of order 2,3,4,5,6,7,10,12. Let a be any element of order mentioned above then you get a cyclic group < a > of order n. Note that there is no element of order 8 in S7. In order to have an element of order 8 in Sn one has break the element into disjoint cycles such that the l.c.m. of the lengths of cycle become 8, which is impossible for S7.

Answer 3

For $n=1,\dots,7$, clearly $S_7$ contains $n$-cycles: each of them is isomorphic to $\mathbb Z_n$. For $n=10$, for example the subgroup $\langle (12)(34567)\rangle$ is isomorphic to $\mathbb Z_{10}$. Likewise, for $n=12$, for example the subgroup $\langle (123)(4567)\rangle$ is isomorphic to $\mathbb Z_{12}$. Can you see now why $S_7$ hasn't got subgroups isomorphic to $\mathbb Z_n$ for $n=8,9,11$?