I am trying to show that for $n\geq 5$, the alternating group $A_n$ has no subgroup of index $p$ where $p$ prime and $p\not = n$. I am supposed to show this without using any of the Sylow theorems.

I understand that $A_n$ is simple for $n\geq5$ but I do not know how to relate this to the existence of subgroups of index $p$. I tried to show this by contradiction, assuming $H \subseteq G$ with $[G:H]=p$ and I managed to show that if this is the case, then $p<n$ but I cannot figure out anything beyond that.

  • $\begingroup$ Well, if you have a subgroup of index $p, you get a homomorphism $A_n\to S_p$ coming from the action of $A_n$ on the cosets $A_n/H$. Can this map be injective? $\endgroup$
    – David Hill
    Mar 9, 2016 at 0:19
  • $\begingroup$ I'm going to hazard a guess no since we could have $agH=bgH$, $a\not=b$. Is that correct? If so, what is the implication? $\endgroup$
    – OnyxDown
    Mar 9, 2016 at 0:57
  • $\begingroup$ It doesn't make sense to avoid using Sylow's Theorem, but to use the more difficult result that $A_n$ is simple for $n \ge 5$. $\endgroup$
    – Derek Holt
    Mar 9, 2016 at 8:08

1 Answer 1


Hint :Let $H$ be a subgroup of index $m>1$ in $A_{n}$, $n \geq 5$. Then, the action of $A_{n}$ on $A_{n}/H$ induces a group homomorphism from $A_{n}$ to $S_{m}$ whose kenel is contained in $H$, and hence is proper! But $A_{n}$ is simple so the map is injective! Hence $\frac{n!}{2} \leq p!$...


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