# Prove a group of order $22,000$ with $16$ Sylow-$5$ subgroup has a normal (Sylow) $11$ subgroup.

Let $$G$$ be a group of order $$2^4\cdot 5^3 \cdot 11$$, $$H$$ be a group of order $$5^3 \cdot 11$$.

1. Prove $$H$$ has a normal $$11$$-subgroup.
2. Suppose $$n_5(G) < 16$$ (number of Sylow $$5$$-subgroups of $$G$$), show that $$G$$ has a normal subgroup such that its order is divisible by $$5$$.
3. Suppose $$n_5(G) = 16$$, use part 1 to prove $$G$$ has a normal $$11$$-subgroup.

I got the first two parts. For the third part, since $$n_5 = 16 = 2^4$$, then if $$S$$ is a Sylow-$$5$$ subgroup of $$G$$ then we have $$n_5 = [G:N_G(S)] = 16$$ thus $$\lvert N_G(S) \rvert = 5^3 \cdot 11$$, thus by part 1, $$N_G(S)$$ has a normal $$11$$-subgroup. If, say, $$N_G(S)$$ is characteristic then we are done, but it does not seem to be true, or at least I don't know how to prove it.

This was on my algebra qualifying exam yesterday. It was the problem I couldn't finish and I'd like to know how to do it... Any hint is appreciated, thanks!

• Since $N_G(S)$ is a Hall subgroup, it suffices to show $N_G(S)$ is normal: see here. Commented Jun 19, 2022 at 22:14

I think you only need elementary Sylow theory to finish the answer of part 3: A subgroup of $$G$$ of order $$11$$ will be a $$11$$-Sylow group. So to show that such a group is normal is equivalent to showing that $$n_{11} = 1$$. You have found a $$11$$-Sylow subgroup and shown that its normalizer contains $$N_G(S)$$. Hence the index of its normalizer must divide $$[G:N_G(S)]=16$$. And since the index of its normalizer must be congruent to $$1$$ modulo $$11$$, you're done.