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Can someone can help me get started with this problems?

Prove that $A_n$ does not have a proper subgroup of index less than $n$ for all $n \geq 5$.

I followed Robert's prove, but I was not able to sort out why it can't be $A_n$. Could someone help me sort this out? Thank you so much!

Tumbleweed
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1 Answers1

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Let $H\leq A_n$. By acting $A_n$ on the cosets of $H$, we get a homomorphism $\Phi:A_n\to\mbox{Bij}(A_n/H)\simeq S_{[A_n:H]}$, where $Bij(X)$ for a set $X$ denotes the group of all bijections $X\to X$ and $A_n/H$ is the set of cosets of $H$. Explicitly, for $x\in A_n$, we get the bijection $A_n/H\to A_n/H$ where $yH\mapsto xyH$. Therefore, we have (by the Isomorphism Theorems) that $$A_n/\ker\Phi\simeq\mbox{image}(\Phi)\leq\mbox{Bij}(A_n/H).$$ Since $A_n$ is simple for $n\geq5$, we have that $\ker\Phi$ is either 1 or $A_n$. It can't be $A_n$, since if $x\notin H$, then $xH\neq H$ and so $\Phi(x)$ is not the identity. Therefore we have that $\Phi$ is an injection and so $|A_n|\mid|\mbox{Bij}(A_n/H)|=|S_{[A_n:H]}|=[A_n:H]!$. If $[A_n:H]<n$, then $n!/2=|A_n|\mid [A_n:H]!<n!$. But $n!/2$ can't divide a number that is strictly less than $n!$, and so we get a contradiction. Therefore $[A_n:H]\geq n$.

rfauffar
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  • Thanks Robert. I don't understand the intuition of acting $A_n$ on the cosets of $H$ - why do we do that? – Tumbleweed Sep 30 '13 at 01:07
  • You're welcome. This is a classic application of group actions. Whenever you have a subgroup you can act on the set of its cosets by left multiplication. Take a look at these notes, for example: http://sierra.nmsu.edu/morandi/notes/groupactions.pdf – rfauffar Sep 30 '13 at 01:14
  • Thanks so much Robert. I read the notes and it is very helpful. I am reading dummit & foote, by any chance do you know any corresponding content in this book? thanks! – Tumbleweed Sep 30 '13 at 01:39
  • I figure my previous question. But I am not quite follow why it can't be $A_n$? – Tumbleweed Sep 30 '13 at 05:28
  • @Tumbleweed He has a fine argument for why it cannot be $A_n$ already there. Which part of that is giving you problems? – Tobias Kildetoft Sep 30 '13 at 08:32
  • @Tumbleweed Take a look at Dummit and Foote chapter 4. Also, the kernel of the morphism can't be all $A_n$ because if it were, then for all $x\in A_n$, we would have that $\Phi(x)=\mbox{id}$. But this means that $xyH=yH$ for all $y\in A_n$, and in particular for $y=1$. But if $x\notin H$, this doesn't happen. – rfauffar Sep 30 '13 at 11:10
  • Hi Robert, I think I got it, so we just pick $x$ not in $H$ and get the result? Thank you! – Tumbleweed Sep 30 '13 at 15:06
  • @Tumbleweed, exactly. – rfauffar Sep 30 '13 at 16:38