Show that $S_n$ has no subgroup of index $t$ for $2 < t < n$

Question

I am trying to solve the following exercise

Show that $S_n$ has no subgroup of index $t$ for $2 < t < n$

which is the exercise 3.25 of An Introduction to Theory of Groups, by Joseph J. Rotman.

I think I should use a theorem that says that if $H \le G$ and $[G : H] = \ell$, then $\exists \rho \in \mathrm{Hom}(G, S_\ell)$ such that $\ker(\rho) \le H$. That would give me an isomorphism between $S_n / \langle \ker(\rho) \rangle $ and $\mathrm{Im}(\rho) \le S_t$ and I was hoping to prove that it is impossible.

I tried to fix some value for $t$, for instance, $t = 3$, to have some idea about the general problem, but even in this scenario, I am not able to prove it, because I can't visualize the group $S_n / \langle \ker(\rho) \rangle $.

Well, do you have any tip? I am not looking for a full answer to that exercise, but maybe you could point out some theorems and properties of $S_n$ that might be useful here.

Answer 1

Using google gave this question here, where someone claimed that he has found a proof: A subgroup of symmetric group of degree n with index smaller than n.

The basic idea is to use the theorem you gave, together with the fact that a kernel of a homomorphism is normal and the only non-trivial normal subgroup of $S_n$ is $A_n$ (for $n \geq 5$, so you might need to do the case $n = 4$ and $t = 3$ by hand).

Answer 2

Theorem 3.14 is used often in solving exercises in Rotman's Introduction to the Theory of Groups. In this application, we are looking for a subgroup $H$ of $G$ such that $t=[G:H]>2$. There is a homomorphism $\rho: G\rightarrow S_t$ with $\text{ker}\;\rho \le H$. $\text{ker}\;\rho$ is a normal subgroup of $S_5$ with an index that is greater than 2. But, $A_5$ is the only non-trivial normal subgroup of $S_5$, so $\text{ker}\;\rho=1$ and $t=n$.