Does $A_{17}$ have a subgroup of index $3$?

Question

I have an exercise which asks me whether the alternating group on $17$ letters $A_{17}$ has a subgroup of index 3.

I found the prime factorisation: $17!/2 = 3^{6}m$ for some integer $m$ coprime to $3$.

Now by Lagrange's Theorem I know that if there exists a subgroup of index $3$ it must have order $3^5m$, but I don't see how this is useful to prove that such a group exists or not.

And by the Sylow theorems I know that I can find a subgroup of order $3^6$. But I also don't know how this is applicable to my exercise.

If anybody could point me on the right path to go about this question, I would appreciate it. :)

Answer 1

Lemma: If $H$ is a proper subgroup of $A_n$ with index $k$ and $n\geq 5$ then $k\geq n$.

Proof: We have a nontrivial homomorphism $\phi$ from $A_n$ to $S_k$ due to the action $A_n$ on the set of distinct left cosets of $H$. On the other hand, $\ker(\phi)$ is trivial, since $A_n$ is simple for $n\geq 5$. Thus, $A_n$ is isomorphic to a subgroup of $S_k$, which yields that $\dfrac{n!}{2}$ divides $k!$ by Lagrange theorem. Clearly, this is not possible when $k<n$ and $n\geq 5$. Thus, $k\geq n$ as desired.

Note that $A_{n-1}$ has index $n$ in $A_n$, and so this index smallest possible in $A_n$ by above lemma.

Answer 2

The group $A_n$ for $n\ge 5$ has no subgroup $H$ of prime index $p\neq n$.

Proof: In fact, the homomorphism from $A_n$ to $S_p$ coming from $A_n$ acting on the cosets of $H$ is injective, since it is not trivial and $A_n$ is simple for $n\ge 5$. Hence by Lagrange $$ \frac{n!}{2} \mid p! $$ However, this can only happen if $n = p$.

In particular, $A_n$ has no subgroup of index $3$ for all $n\ge 5$.