Show that |G|=168 has 8 sylow 7 subgroups and find the normalizer of one of these.

Question

I have shown that there are 8 sylow subgroups, but i dont really know where to start with finding the normalizer.

p=7 => n_7={factors of 24}={1,2,3,4,6,8,12,24}congruent 1modp

so n_7=1 or 8, but it cannot be 1, hence n_7=8 as required.

Answer 1

Note: I assume you are talking about a simple group of order 168, as in your earlier question.

Let $P$ be a 7-Sylow subgroup. We know that $|N_G(P)| = 168/8 =21$. There are only 2 groups of order 21 - a cyclic group and a non-abelian group. I claim that $N_G(P)$ is the non-abelian group.

To see this, it suffices to prove that $G$ does not have an element of order 21. But first, we need a Lemma :

Lemma : Let $n_3$ denote the number of 3-Sylow subgroups, then $7\mid n_3$

Proof :

We know that $n_3 \mid 56$, so if $7\nmid n_3$, then $n_3\mid 8$.

Since $n_3 \equiv 1\pmod{3}$, and $n_3\neq 1$, the only possibility is that $n_3 = 4$. Now let $G$ act on the 3-Sylow subgroups by conjugation to give a non-trivial homomorphism $G\to S_4$. Since $|G| \nmid 24$, it follows that the kernel of this homomorphism would be a non-trivial normal subgroup. This is impossible, so $n_3 \neq 4$.

Hence, $7\mid n_3$.

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Proposition : $G$ does not have an element of order 21.

Proof :

Suppose $x\in G$ has order 21, then $Q:= \langle x^7 \rangle$ is a 3-Sylow subgroup, and $x^3 \in N_G(Q)$. But $|x^3| = 7$, so $7\mid |N_G(Q)|$, whence $7\nmid n_3 = [G:N_G(Q)]$. This contradicts the previous lemma.

Hence, there does not exist an element of order 21 in $G$; and so $N_G(P)$ is the unique non-abelian group of order 21.