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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!

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

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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.

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