Let $G$ be a group of order $2016 = 2^5 \cdot 3^2 \cdot 7$ in which all elements of order $7$ are conjugate.

Question

Let $G$ be a group of order $2016 = 2^5 \cdot 3^2 \cdot 7$ in which all elements of order $7$ are conjugate. Prove that $G$ has a normal subgroup of index $2$

I know any subgroup of index $2$ must be normal, and I feel like the way the question is worded I should start off by letting $\Omega$ be the set of elements of $G$ of order $7$ and then let $G$ act on $\Omega$ by conjugation and then get that we have a homomorphism from $G$ into $S_{|\Omega|}$ and let $K$ be the kernel of this action. So $K$ is a normal subgroup of $G$.

My lack of intuition tells me that I should try to prove $k$ is the group of index $2$ I am looking for, and that I should some how use the fact that $A_{|\Omega|}$ is a normal subgroup of $S_{|\Omega|}$ of index $2$ to do this, but I can't figure out.

I feel like it would be helpful if I could figure out the order of $\Omega$. By Sylows Theorem I know that there are $1, 8, 36$, or $288$ subgroups of order $7$ so the order of $\Omega$ must be $6, 48, 216$, or $1728$.

Also the fact that for any subgroup $H < G$ we have that $|G:H|$ divides $7!$ iff $2$ divides $H$ has very much got my attention but don't see how its relevant.

Any help would be appreciated.

Answer 1

Here are some ideas. I will leave you to fill in the details. Of course there might be easier ways to do it!

The 6 nontrivial elements in a Sylow $7$-subgroup $P$ are all conjugate, and they must be conjugate in $N_G(P)$, so $|N_G(P)|$ is divisible by $6$. That rules out $36$ and $288$ for the number of Sylow $7$-subgroups.

If there is a single Sylow $7$-subgroup, then $P \lhd G$, and also $C_G(P) \lhd G$ with $|G/C_G(P)|=6$. So $G/C_G(P)$ and hence also $G$ have subgroups of index 2.

When there are $8$ Sylow $7$-subgroups, the image of the conjugation action of $G$ on $\Omega = {\rm Syl}_7(G)$ is a 3-transitive group of order $8 \times 7 \times 6$, where the 2-point stabilizer contains a $6$-cycle, which is an odd permutation. So intersecting with ${\rm Alt}(\Omega)$ gives a subgroup of index 2.