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It seems like proving that groups of a certain size are never simple is usually done with Sylow theorems, showing that a Sylow subgroup of a particular size must be normal. But is there an example group size $n$ where no groups of size $n$ are simple, yet there is a group of size $n$ which has no normal Sylow subgroups? If so, how does the proof go for that size $n$ that no groups of size $n$ are simple?

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up vote 5 down vote accepted

The smallest example is $n = 24$.

The symmetric group $S_4$ is a group of order $24$ with no normal Sylow subgroups, and there are no simple groups of order $24$.

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Does the proof of the first statement consist of checking cases? Or is there a way more elegant? – RghtHndSd Apr 6 '14 at 1:09
@rghthndsd: Checking cases is not so bad here. First of all, a prime power $p^k$ cannot be an example, and neither can $pq$ for $p$, $q$ primes. That already counts out everything except $12, 18$ and $20$. You can see immediately that a group of order $18$ has a normal $3$-Sylow, and a group of order $20$ has a normal $5$-Sylow. So that leaves you with proving that a group of order $12$ has a normal Sylow subgroup. That can be proven with a counting argument (if a $3$-Sylow is not normal, then by counting elements you see that a $2$-Sylow must be normal.). – Mikko Korhonen Apr 6 '14 at 8:08
Bonus brownie points and accepted answer for giving the smallest possible example. Thanks! It'd be even better if you could show the proof that there are no simple groups of size $24$, although I believe it because it seems like either $G = S_4$ or $G$ has a normal Sylow subgroup. – user2566092 Apr 6 '14 at 19:37
@user2566092: yes, by considering the conjugation action on $3$-Sylows you can prove that $G \cong S_4$ if $G$ has order $24$ and has no normal Sylow subgroups. – Mikko Korhonen Apr 6 '14 at 19:58

I would suggest that the symmetric group $S_5$ of order $120$ might fit the bill. It is not simple, but the only non-trivial normal subgroup is $A_5$ of order $60$ which is not a Sylow subgroup.

There are no simple groups of order $120$.

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Sketch of proof. Suppose a group of order $120$ is simple. It has Sylow subgroups of order $8$, $3$ and $5$. The number of subgroups of order $5$ must be $6$, and these are permuted transitively by conjugation. This identifies the group with a subgroup of $S_6$, which can be identified with $S_5$. $S_5$ is not simple. The embedding of $S_5$ in $S_6$ in this way is of great interest in other contexts. – Mark Bennet Apr 5 '14 at 21:20

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