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How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?

Why isn't there other possibility?

Thanks :)

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Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening. –  Mariano Suárez-Alvarez Mar 14 '11 at 21:00
    
I am terribly sorry. But I don't know how to reedit it. –  ShinyaSakai Mar 14 '11 at 21:07
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There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit. –  Arturo Magidin Mar 14 '11 at 21:11

2 Answers 2

up vote 13 down vote accepted

As mentioned by yoyo: if $H\subset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2=\{1,-1\}$. We thus have a surjective homomorphism $f:S_n\to C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)\in C_2$ is the same element for every transposition $t\in S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $t\in S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.

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Very detailed~ thank you very much~ –  ShinyaSakai Mar 15 '11 at 21:28
    
$S_n$ in the last line, not $S_2$. Very nice solution. –  Genomeme Sep 7 at 23:41
    
@Genomeme: thanks, corrected –  user8268 Sep 8 at 21:14

subgroups of index two are normal (exercise). $A_n$ is simple, $n\geq 5$ (exercise). if there were another subgroup $H$ of index two, then $H\cap A_n$ would be normal in $A_n$, contradiction.

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It is really a smart shortcut for the special case of $n \geq 5$~ Thank you very much~ –  ShinyaSakai Mar 15 '11 at 21:30
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How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$...... –  Vladhagen Oct 30 '13 at 19:42

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