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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?

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  • $\begingroup$ What is $G$ and the embedding theorem? $\endgroup$
    – Hanul Jeon
    Commented May 15, 2015 at 13:49
  • $\begingroup$ Which embedding theorem? Skorokhod, Nash, Gabriel-Popescu, Freyd-Mitchell, Whitney, Sobolev, Kodeira, Higman, Hahn, Campbell, Assouad, ... ? $\endgroup$ Commented May 15, 2015 at 13:50
  • $\begingroup$ 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 $A_n$. $\endgroup$
    – Jasmine
    Commented May 15, 2015 at 14:04
  • $\begingroup$ 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 $A_n$? $\endgroup$
    – Nishant
    Commented May 15, 2015 at 15:20

2 Answers 2

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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 $n$.

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  • $\begingroup$ Can you please explain how you reached from kernel being $A_n$ to $G =$ $A_n$? $\endgroup$
    – Jasmine
    Commented May 15, 2015 at 14:12
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    $\begingroup$ 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 $G=A_n$. $\endgroup$
    – Nishant
    Commented May 15, 2015 at 14:19
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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.

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    $\begingroup$ Hi Shreya. Welcome to MSE! It is helpful to use MathJax formatting to make your answers more readable. $\endgroup$
    – eepperly16
    Commented Dec 1, 2017 at 7:06
  • $\begingroup$ Also I've taken the liberty of editing your post to add MathJax and change your language to hopefully be more clear. If you don't like what I've done, feel free to change it back. $\endgroup$
    – eepperly16
    Commented Dec 1, 2017 at 7:09
  • $\begingroup$ From embedding theorem we obtain a stronger condition, that, $G$ must be $\simeq$ $A_n$ $\endgroup$
    – So Lo
    Commented Dec 11, 2018 at 15:20

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