Describing the normalizer of a cycle in $S_{10}$

Question

Let $\sigma = (1,2,3,...,9) \in S_{10}\;$ a $10$-cycle in the permutation group $S_{10}$

a. What is the size of $N_{S_{10}}(\langle \sigma \rangle)$, The normalizer of the group generated by $\sigma$ ?

b. Describe what are the permutaions in $N_{S_{10}}(\langle \sigma \rangle)$.

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My attempt for (a): $S_{10}$ acts on its set of subgroups by conjugation. We have: $$O(\langle \sigma \rangle)=\{g\langle \sigma \rangle g^{-1} | g \in S_{10} \}=\{\langle g \sigma g^{-1} \rangle | g \in S_{10} \}$$

Since $g \sigma g^{-1}$ has the same cycle structure of $\sigma$, $O(\langle \sigma \rangle)$ is the set of all cyclic subgroups generated by a $9$ cycle. The number of permutations with this cycle structre is $\frac{10!}{9}$. Since there are $\phi(9)=6$ generators for each cycle subgroup generated by a $9$-cycle, we have $|O(\langle \sigma \rangle)|$ = $\frac{10!}{6 \cdot 9}$. Now we have $$|N_{S_{10}}(\langle \sigma \rangle)|= \frac{|S_{10}|}{|O(\langle \sigma \rangle)|}=\frac{10!}{\frac{10!}{54}}=54$$

Is this correct? also, I would like to hear ideas for (b).

Answer 1

Part a is spot on.

As to part b, note that $C_{S_{10}}(\langle \sigma \rangle)$ has order $9$ by orbit-stabiliser, and thus it coincides with $\langle \sigma \rangle$. Also, $54 = 9 \cdot 6$, and $54/9 = 6 = \varphi(9)$ is the size of the automorphism group of the cyclic group of order $9$ like $\langle \sigma \rangle$.

Spoiler 1

The automorphism group of $\langle \sigma \rangle$ is generated by $f : \sigma \mapsto \sigma^{2}$.

Spoiler 2

The permutation $\tau = (2, 3, 5, 9, 8, 6) (4, 7)$ conjugates $\sigma$ to $\sigma^{2} = (1, 3, 5, 7, 9, 2, 4, 6, 8)$.