Find the normalizer of a cyclic subgroup of $S_7$

Question

Let $P\subset S_7$ be a cyclic subgroup of order $7$. Show that the normalizer $N$ of $P$ has order $42$, and find a pair of cycles generating $N$.

My attempt

First note that the cyclic subgroup is a Sylow-7 subgroup of $S_7$. Let the number of Sylow-7 subgroups in $S_7$ be denoted $n_7$.

The Sylow-7 subgroups all intersect trivially and are all cyclic, so every nonidentity element in a Sylow-7 subgroup is of order $7$ and the number of elements in $S_7$ with order $7$ is equal to $n_7$ times $6$.

The elements of order $7$ in $S_7$ are the 7-cycles. And there are $7!/7$ of them.

Therefore, $n_7 = 5!$.

Now, to show that $|N|=42$, use the identity: $n_7 = |S_7 : N| = |S_7|/|N|$.

My question

First, I can't see why $N$ is generated by a pair of cycles.

Also, I am having trouble writing out explicitly the generators of $N$.

Any help would be greatly appreciated.

Answer 1

Hint: As you noticed, the elements of order $7$ are the $7$-cycles. WLOG, let $P$ be the group generated by $(1\ 2\ 3 \ 4 \ 5 \ 6\ 7) = \sigma$. Then clearly $P$ normalizes itself.

If we use the fact than in any group $G$, for two subsets $H,K$ of $G$, we have $$|HK| = \frac{|H| |K|}{|H \cap K|} $$ we can form a set of order $42$ if we let $H = P$ and $K$ be a group generated by a $6$-cycle. If this $6$-cycle happens to normalize $P$, we've formed a set of order $42$ where each element normalizes $P$. Since $|N_G(P)| = 42$, this set must be $N_G(P)$. So you just need to find a $6$ cycle that normalizes $P$

I will throw in the additional fact that for any $k$-cycle $(a_1\ a_2\ ... a_k) = x$ and permutation $\tau$, we have that $\tau x \tau^{-1} = (\tau (a_1)\ \tau( a_2) \ ... \tau( a_k))$