# Proving a theorem about a finite simple group

I need to prove this theorem that I had in an exam and I am stuck.

Let $G$ be a finite simple group and we assume that for every prime $p$ the number of $p$-sylow sub-groups is $\leq 6$. Prove that $G$ is cyclic.

my lead: I proved that $|G| = 2^a3^b5^c$ (no other primes divides the size of $G$). and now I've thought that maybe I can prove that $a=1$ or $b=1$ or $c=1$ and continue from there (I've proved before that

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You'll in fact have to show that atleast two of $a,b,c$ is $0$. Because, only simple cyclic groups are those of prime order. – user21436 Feb 18 '12 at 21:15
Can you please add the proof of your claim about the order of the group? Something in me tells you're claim is wrong. For instance, take a group of order $23$. It satisfies all your hypothesis, but your claim does not reflect this. – user21436 Feb 18 '12 at 21:25
I did the same proof that m. k. did – benjamin Feb 19 '12 at 10:06

We can assume that $G$ is not trivial.

If $G$ has a normal $p$-Sylow subgroup, then $G$ must have order $p^k$ since $G$ is simple. Groups of prime power order have a nontrivial center, so $G$ has order $p$ and is thus cyclic.

Suppose then that no Sylow subgroup of $G$ is normal. We will show that this leads into a contradiction. Now for each prime divisor $p$ of $G$ there are $2$, $3$, $4$, $5$ or $6$ $p$-Sylow subgroups. The amount of $p$-Sylow subgroups is $\equiv 1 \mod p$, so the only prime divisors of $G$ can be $2$, $3$ and $5$.

If $G$ has order divisible by $3$, then there are four $3$-Sylow subgroups. Therefore there exists a homomorphism $\phi: G \rightarrow S_4$ with $\operatorname{Ker}(\phi)$ contained in the normalizer of a Sylow $3$-subgroup. Since $G$ is simple, the kernel is trivial and thus $G$ is isomorphic to a subgroup of $S_4$. Now $S_4$ has order $2^3 \cdot 3$, so $G$ has three $2$-Sylow subgroups. Thus we can embed $G$ into $S_3$, but this isn't actually possible when $G$ has four $3$-Sylow subgroups.

Therefore the order of $G$ is $2^a5^b$ for some integers $a$ and $b$. If $b > 0$, then $G$ would have six $5$-Sylow subgroups. This is not possible, because $6$ does not divide $2^a5^b$. Thus $G$ has order $2^a$ and must be cyclic, contradicting the assumption that $G$ has no normal Sylow subgroups.

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A group of order $2^a5^b$ cannot have $6$ Sylow $5$-subgroups, because $6$ doesn't divide $|G|$! – user641 Feb 18 '12 at 22:15
@SteveD: Thanks, that is much better. I edited my answer. – Mikko Korhonen Feb 18 '12 at 22:21
thanks for the answer but I don't understand the second paragraph. if G is of order 3^k it can't have four 3-sylow subgroups because four doesn't divide 1. and else I don't understand why G is of prime order. – benjamin Feb 18 '12 at 23:36
So it can't have four Sylow 3-subgroups, so that's a contradiction. So $G \le S_4$. Now $G$ must have 3 Sylow 2-subgroups, so $G \le S_3$, etc. – Derek Holt Feb 19 '12 at 4:34
why is it a contradiction? if a>1 then there can be four 3-sylow subgroups – benjamin Feb 19 '12 at 9:30