Normalizer of the cyclic group in $S_n$

Question

Let $G = S_n$ and $H = \langle (1,2,\ldots,n) \rangle.$ It is not too hard to see that $$C_G(H) = H.$$

What I am now wondering is, which group is $N_G(H)?$ Is there any way to determine that?

I guess the key is to use the fact that $$g (1,2,\ldots, n) g^{-1} = (g(1), \ldots, g(n))$$ but I don't see how to get the full answer just by this.

Answer 1

Consider the following bits/steps. I abbreviate $\alpha=(123\ldots n)$.

• If $g\in N_G(H)$, then $g\alpha g^{-1}\in H$, and $g\alpha g^{-1}$ must also have order $n$.
• Therefore $g\alpha g^{-1}=\alpha^k$ for some integer $k$, $\gcd(k,n)=1$.
• All the elements $\alpha^k$, $k$ as above, are conjugate in $S_n$.
• The elements $g$ such that $g\alpha g^{-1}=\alpha^k$, $k$ as above, form a coset of $C_G(H)=H$.
• The number of such cosets in $N_G(H)$ is thus given by the Euler totient function $\phi(n)$.
• $N_G(H)$ will be isomorphic to $C_n\rtimes Aut(C_n)$.

Answer 2

Perhaps it is worth noting the more general fact, that if $H$ is a group of order $n$, and $\psi : H \to S_{n}$ is the regular representation, then $N_{S_{n}}(\psi(H))$ is isomorphic to the holomorph $H \rtimes \operatorname{Aut}(H)$ of $H$.