Groups with 20 Sylow subgroups 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).
 A: 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.
