Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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

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 $n\geq 5$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.