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Problem

Let $G$ be a group of order $2^3 7^2 11$.

a) Show $G$ has a subgroup of order 77.

b) Show there there exists a subgroup $Q$ of order 7 such that 8 divides the index of the normalizer of $Q$.

My Progress

a) I have solved part a:

Sylow's Theorem tells us that the number of Sylow 11 subgroups, denoted $n_{11}$ satisfies:

(i) $n_{11} = 1$ (mod11)

(ii) $n_{11} / 2^3 7^2$

So $n_{11} = 1$ or $56.$

Case 1: n_{11} = 1. Then the Sylow-11 subgroup $P$ is normal. Sylow's theorems also tells us that if $p$ is a prime that divides the order of $G$, then there is a subgroup of that order. So there exists a subgroup $R$ of order 7. The direct product of a normal subgroup and a subgroup is a subgroup. So $PR$ is a subgroup of order: $|PR|= \dfrac{|P||R|}{|P \cap Q|}= \dfrac{77}{1}=77.$

Case 2: n_{11}=56. Let $P$ denote a sylow-11 subgroup. Recall that $n_{11}= |G: N_G(P)|$. It follows that $N_G(P)$ is a subgroup of order 77.

b) Firstly, there is a mistake in the statement of the question. If we don't assume that $G$ is not Abelian, then the index of the normalizer of any subgroup is 1, and clearly 8 does not divide 1.

So let us assume that $G$ is not Abelian.

Any ideas what to do from here?

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  • $\begingroup$ Please ask one question at a time. $\endgroup$ – Shaun Mar 24 '18 at 19:09
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    $\begingroup$ My question is how to solve part b. I just provided what I had for part a because it is likely useful in solving part b. $\endgroup$ – Squeeze my theorem Mar 24 '18 at 19:19
  • $\begingroup$ Fair enough. ${}{}$ $\endgroup$ – Shaun Mar 24 '18 at 19:21

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