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Is there a reasonably easy proof that a finite group with exactly $20$ Sylow $p$-subgroups has PSL$(2,19)$ or PGL$(2,19)$ as a quotient group?

What if we weaken this to merely: “a group of order $760$ has a normal Sylow 19-subgroup”?

One can see this How to show there are no simple groups of order 760 using Sylow's theorem for some motivation. My motivation is merely to turn this into a more positive statement, but the arguments for 760 are getting pretty complicated (one can easily show a group of order 760 either (1) has a normal subgroup of index 2, (2) a normal subgroup of size 2 and a normal subgroup of index 19, or (3) a normal subgroup of size 19; however, every group in fact lands in case (3) through a convoluted theoretical argument or a quick check of computer databases).

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    $\begingroup$ OK, preliminary step: Since the number of $p$-Sylow groups is $\equiv 1\pmod p$, the number of $20$ Sylow groups immediatly gives us that $p=19$. But if $G$ has 20 Sylow $19$-groups, doesn't $G\times A$ as well as long as $|A|$ is not a multiple of $19$? $\endgroup$ – Hagen von Eitzen Apr 23 '13 at 20:04
  • $\begingroup$ @HagenvonEitzen: good point. $G/Core(N_G(P))$ is isomorphic to one of those two groups, but $G$ itself can be bigger. $\endgroup$ – Jack Schmidt Apr 23 '13 at 20:08
  • $\begingroup$ (For Jack's reference: math.stackexchange.com/questions/18467/… describes the core in the p-core free case) $\endgroup$ – Jack Schmidt Apr 23 '13 at 20:11
  • $\begingroup$ How do you reduce to cases (1), (2) and (3)? $\endgroup$ – Mikko Korhonen Apr 23 '13 at 21:56
  • $\begingroup$ I wonder if you can use the conjugation action of $G$ on the Sylows to, WLOG, embed $G$ into $S_{20}$ as a doubly transitive group. The $760$ case shows there must be some elements of order $3$. Basically, the odd order subgroup of $N_G(P)$ in $S_{20}$ "should" give the subgroup of order $19\cdot9$ in $PSL(2,19)$. I have no idea how to prove $G$ has a subgroup of order $20$. $\endgroup$ – user641 Apr 23 '13 at 23:49
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Suppose that $G$ is of order $760 = 2^3 \cdot 5 \cdot 19$. Denote by $n_p$ the number of Sylow $p$-subgroups in $G$.

We can prove that $G$ has a normal Sylow $5$-subgroup and a normal Sylow $19$-subgroup. To do this, note first that it suffices to prove the existence of a subgroup $H$ of order $95 = 5 \cdot 19$. Then $H$ normalizes a $5$-Sylow and a $19$-Sylow. Hence if $P$ is a $5$-Sylow or a $19$-Sylow, then $|N_G(P)|$ will be divisible by $95$ and $[G : N_G(P)]$ is a power of $2$. By Sylow's theorem, $[G : N_G(P)] = 1$. Thus we only need to prove that a Sylow $5$-subgroup or a Sylow $19$-subgroup is normal in $G$.

So let's assume that Sylow $5$- and $19$-subgroups are not normal in $G$. Then $n_5 = 76$ and $n_{19} = 20$. Thus Sylow $5$-subgroups have a normalizer of size $10$, and Sylow $19$-subgroups have a normalizer of size $38$. Note that this implies that $G$ cannot contain a subgroup of order $2^k \cdot 5$ or $2^k \cdot 19$ for $k \geq 2$. Such a subgroup would normalize a $5$-Sylow or a $19$-Sylow, making a normalizer too big. This shows that the cases $n_2 = 1$, $n_2 = 5$ and $n_2 = 19$ are not possible.

We are left with the case $n_2 = 95$. Now $n_2 \not\equiv 1 \mod{4}$, so there exist two Sylow $2$-subgroups that intersect in a subgroup $D$ of order $2^2$. Then $|N_G(D)|$ is a multiple of $2^3$ and larger than $2^3$, so $|N_G(D)| = |G|$ because $|N_G(D)| = 2^3 \cdot 5$ and $|N_G(D)| = 2^3 \cdot 19$ are not possible. Thus $D$ is a normal subgroup, which gives us the existence of a subgroup of order $20$, a contradiction.

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  • $\begingroup$ You can do this without caring about the Sylow 5- or 19-subgroups. If there are 95 Sylow 2-subgroups, there is a subgroup of index 5 or 19, which then contains a normal Sylow 19- or 5-subgroup. Sylow's theorems show this is normal in all of G. Other counts for Sylow 2-subgroups reduce once again to this case. $\endgroup$ – user641 Apr 23 '13 at 23:12
  • $\begingroup$ @SteveD: That would simplify the proof. How do you find the subgroup of index $5$ or $19$ when $n_2 = 95$? $\endgroup$ – Mikko Korhonen Apr 24 '13 at 16:08
  • $\begingroup$ @m.k. I'm not positive this simplifies the proof, but a group whose order is not divisible by 16 or 3 or 7 is always 2-nilpotent, so groups of order 760 have a normal Hall {5,19}-subgroup, which must be abelian as in your argument. The 2-nilpotency follows from Frobenius's p-complement theorem since an odd order automorphism of a group of order dividing 8 must have order 3 or 7. $\endgroup$ – Jack Schmidt Apr 25 '13 at 14:34

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