# Prove that $A_6$ has no subgroup of prime index

This is an exercise in Rotman's Advanced Modern Algebra. Keep in mind no Sylow subgroups theory is developed. Only until group actions.

So $p=2$ is fine, since $A_{6}$ is simple. Suppose that $H$ is a subgroup of index $3$. Consider the action on the set of subgroups of $A_{6}$. Since the order of the orbit of $H$ is equal to $[A_{6}:N_{A_{6}}(H)]$ and $3=[A_{6}:H]=[A_{6}:N_{A_{6}}(H)][N_{A_{6}}(H):H]$, and the action on the set of subgroups of $A_{6}$ cannot be transitive, otherwise $H$ is normal, I get that $H$ is equal to its normalizer. But I don't know much about the normalizer, and fiddling around I don't seem to get a contradiction. I don't know whether I should be using the order of $H$, maybe?

Any hint or help is appreciated. Thanks in advance!

• In general, if a simple group $G$ has a subgroup of index $n$ then $|G|$ divides $n!$. And of course we also have that $n$ divides $|G|$. This rules out all primes for $A_6$. – Tobias Kildetoft Jan 22 '14 at 13:51
• Just kidding :) looked at the subgroup structure and I don't see any prime index groupprops.subwiki.org/wiki/… – ir7 Jan 22 '14 at 13:58
• Note that my comment actually shows the same statement for any simple alternating group. – Tobias Kildetoft Jan 22 '14 at 14:00
• Thanks all. I've got $\phi : A_{6}\rightarrow S_{3}$ where $\phi (\alpha)$ is the permutation representing the action on the left cosets $A_{6}/H$. The kernel is normal so either trivial or $A_{6}$ itself. Both give contradictions. I can see that generalizing this gives that no subgroup of index less than $\frac{1}{2}n$ exists in $A_{n}$. I think in the same way it can prove your statement, Tobias (because then G will be isomorphic to a subgroup of $S_{n}$?). I'll accept it as an answer if you want to write it. Thanks again :) – user68193 Jan 22 '14 at 14:44
• There are 144 5-cycles in A6, and the hypothetical H is of order 120. There are 3 copies of H... – Richard Peterson Jan 22 '14 at 14:45

In general, if $G$ is a simple group and $H$ is a subgroup of $G$ of index $m > 1$, then the homomorphism from $G$ to $S_m$ coming from $G$ acting on the cosets of $H$ will be injective (since it is not trivial and $G$ is simple).
Thus, we get that $n$ divides $|G|$ by Lagrange, and that $|G|$ divides $m!$ (also by Lagrange).
This means that if $A_n$ has a subgroup of index $p$ for some prime $p$ (and for $n\geq 5$), then $\frac{n!}{2}$ divides $p!$, which can only happen if $n = p$. On the other hand, as pointed out by Jack Schmidt in the comments, $A_{p-1}$ does indeed have index $p$ in $A_p$.