Alternating Group $A_n$ does not have proper subgroup of index less than n, where n>4.

A proof is to be given for this. So what i have thought is: Let us assume to the contrary, i.e. it does have a subgroup of index m (say) less than n. Then, since $A_n$ is simple for n>4 , by embedding theorem, $A_n$ is isomorphic to a subgroup of $A_m$. but this is not possible as the order of $A_n$ (n!/2) does not divide the order of $A_m$ (m!/2) {since $m<n$}. Is this reasoning correct?

• What is $G$ and the embedding theorem? Commented May 15, 2015 at 13:49
• Which embedding theorem? Skorokhod, Nash, Gabriel-Popescu, Freyd-Mitchell, Whitney, Sobolev, Kodeira, Higman, Hahn, Campbell, Assouad, ... ? Commented May 15, 2015 at 13:50
• G is $A_n$ and Cayley's embedding theorem. It states that if G is a finite non-Abelian simple group and H is a subgroup of index n, then G is ismorphic to a subgroup of $A_n$. Commented May 15, 2015 at 14:04
• Wait, I can see how it follows that $G$ is isomorphic to a subgroup of $S_n$, but how do you show it's a subgroup of $A_n$? Commented May 15, 2015 at 15:20

Let's assume $n\geq 5$, since $C_3\leq A_3$ has index $2$ and $V_4\leq A_4$ has index $3$.
Assume $A_n$ has a subgroup $G$ of index $m<n$. Then the action on the cosets of $G$ gives a homomorphism into $S_m$. Since $n\geq 5$, $n!/2>m!$, so the homomorphism can't be injective. Since $A_n$ is simple, the kernel must be all of $A_n$. In particular, this means that $hG=G$ for all $h\in A_n$, which is only possible if $G=A_n$, and is thus improper. Thus, there is no proper subgroup of index less than $n$.
• Can you please explain how you reached from kernel being $A_n$ to $G =$ $A_n$? Commented May 15, 2015 at 14:12
• The action on cosets is given by $h\cdot kG=(hk)G$. If $h$ is in the kernel of the homomorphism, then $hkG=kG$ for all $k$. In particular, if we set $k=1$, we get $hG=G$ for all $h$ (since we have that everything is in the kernel). This means there is only one coset of $G$ in $A_n$, which means $G=A_n$. Commented May 15, 2015 at 14:19
Assume $A_n$ has a subgroup of index $m < n$. Since $n>4$, $n!/2 > m$. Then, by the Embedding Theorem, which states that, if $G$ is a finite non-Abelian simple group and $H$ is a subgroup of index $n$, then $G$ is isomorphic to a subgroup of $S_n$, we get that $A_n$ is isomorphic to a subgroup of $S_m$, that is, $n!/2$ divides $m!$, which is a contradiction. Hence our assumption was false.
• From embedding theorem we obtain a stronger condition, that, $G$ must be $\simeq$ $A_n$ Commented Dec 11, 2018 at 15:20