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I tried so much to prove the following fact about the alternating groups $A_n$, $n\geq 5$ but I couldn't prove it. Any answer or hint will be appreciate;

Any maximal subgroup of the mentioned groups has size greater than n.

Many thanks!

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Do you already know $\,A_n\,\,,\,n\geq 5\,$ , is simple? –  DonAntonio Jan 24 '13 at 12:18
Yes of course! Is this property solve the problem? –  shankfei Jan 24 '13 at 12:28
It might help to think about why this fails for $2\leq n\leq 4$. –  Colin McQuillan Jan 24 '13 at 13:13

1 Answer 1

Consider a maximal subgroup $H$ of $\mathrm{Alt}(\{1,\dots,n\})$. The orbits $X_1,\dots,X_k$ of $H$ form a partition of $\{1,\dots,n\}$. If there is only one orbit ($H$ is transitive) then $|H|\geq |X_1|=n$. Otherwise $H$ is contained in $H'=\mathrm{Alt}(\{1,\dots,n\})\cap(\mathrm{Sym}(X_1)\times\mathrm{Sym}(X_2\cup\dots\cup X_k))$, and by maximality of $H$ we have $H=H'$. So $|H|=|H'|=|X_1|! (n-|X_1|)!/2$, which is at least $n$ for $n\geq 5$.

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The question actually asks for greater than $n$, not greater than or equal. Since $k!(n-k)!>2n$ for $n\geq 5$ and $0<k<n$, you still need to prove (only) that an $n$-cyclic subgroup of $A_n$ is not maximal (if $n$ is odd). –  Marc van Leeuwen Jan 24 '13 at 13:14
@MarcvanLeeuwen: good point! I will leave that part as an exercise. :-) –  Colin McQuillan Jan 24 '13 at 13:18
@ Colin McQuillan Thank you so much. –  shankfei Jan 24 '13 at 13:56
Surely you need to prove that an arbitrary transitive subgroup of order $n$ (that is, a regular subgroup) is not maximal. –  Derek Holt Jan 24 '13 at 14:04
@ Derek Holt thanks. –  shankfei Jan 24 '13 at 14:22

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