Reference request on centralizer of symmetric group

Question

I am trying to understand eq. 2.4 of the paper: https://arxiv.org/pdf/hep-th/9608096.pdf . The statement is:

The centralizer subgroup of a permutation g in this conjugacy class takes the form $C_g = S_{N_1} × (S_{N_2} \rtimes Z_2^{N_2} ) × . . . (S_{N_s} \rtimes Z_s^{N_s} ).$

(the conjugacy class is expressed as $[g]=(1)^{N_1}(2)^{N_2}....(s)^{N_s}.$) I am looking for a proof of the formula. Also, intuitively how wreath product enters to the formula isn't clear to me.

Answer 1

The basic idea of the proof is this. Set $\Omega = \{1,2,\ldots,n\}$.

• You can write $\Omega$ as a disjoint union $\Omega = \Omega_1 \cup \Omega_2 \cup \cdots$, where $g \Omega_i = \Omega_i$ and $g$ acts on $\Omega_i$ as a product of $N_i$ cycles of length $i$.
• If $S \subseteq \Omega$ is $g$-invariant and $z \in C_g$, then $zS$ is also $g$-invariant. Conclude that $z\Omega_i \subseteq \Omega_i$ for all $z \in C_g$ and $i$.
• So $C_g = C_{Sym(\Omega_1)}(g_1) \times C_{Sym(\Omega_2)}(g_2) \times \cdots$ where $g_i$ is the image of $g$ in $Sym(\Omega_i)$.
• What you still have to show that for $g_i \in Sym(\Omega_i)$ a product of $N_i$ cycles of length $i$, the centralizer is $S_{N_i} \wr \mathbb{Z}/i\mathbb{Z} = S_{N_i} \ltimes (\mathbb{Z}/i\mathbb{Z})^{N_i}$. I will leave the details to you, but here $(\mathbb{Z}/i\mathbb{Z})^{N_i}$ is the product of the cyclic subgroups generated by the disjoint cycles of $g$, and $S_{N_i}$ acts by permuting the cycles.