## Compute the centralizer of a product of two 2-cycles in Sym(n)

Show that $|C_{S_n}((1\ 2)(3\ 4))| = 8 \cdot (n-4)!$ for all $n \geq 4$. Determine the elements in this centralizer explicitly.

First we prove a lemma.

Lemma: If $x \in G$ has order 2, then $C_G(x) = N_G(x)$. Proof: $(\subseteq)$ If $g \in C_G(x)$, then $gx = xg$. Then $g \langle x \rangle = \{ g, gx \} = \{ g, xg \} = \langle x \rangle g$. Thus $g \in N_G(x)$. $(\supseteq)$ If $g \in N_G(x)$, then $\{ g, gx \} = g \langle x \rangle = \langle x \rangle g = \{ g, xg \}$. If $gx = g$, then $x = 1$, a contradiction. So $gx = xg$, hence $g \in C_G(x)$. $\square$

We know that the number of conjugates of $(1\ 2)(3\ 4)$ in $S_n$ for $n \geq 4$ is the number of products of disjoint 2-cycles. In this previous exercise we found that this number is $n(n-1)(n-2)(n-3)/8$.

Thus $[S_n : N_G((1\ 2)(3\ 4))] = n(n-1)(n-2)(n-3)/8$; using Lagrange’s Theorem and since $S_n = n!$, we have $|N_G((1\ 2)(3\ 4))| = 8 \cdot (n-4)!$. By the Lemma, $|C_G((1\ 2)(3\ 4))| = 8 \cdot (n-4)!$.

Now all the permutations which fix $1,2,3,4$ centralize $(1\ 2)(3\ 4)$, and there are $(n-4)!$ of these. In addition, $(1\ 2)$ and $(3\ 4)$ centralize $(1\ 2)(3\ 4)$. Thus all elements of the form $(1\ 2)^a(3\ 4)^b\tau$ where $\tau$ fixes $1,2,3,4$ centralize $(1\ 2)(3\ 4)$ and $0 \leq a,b \leq 1$ (not both 0); there are $3 \cdot (n-4)!$ such elements. Moreover, note that $(1\ 3)(2\ 4)(1\ 2)(3\ 4) = (1\ 4)(2\ 3)$ and $(1\ 2)(3\ 4)(1\ 3)(2\ 4) = (1\ 4)(2\ 3)$, so that $(1\ 3)(2\ 4)$ centralizes $(1\ 2)(3\ 4)$. Thus every element of the form $(1\ 3)(2\ 4)\tau$, where $\tau$ fixes $1,2,3,4$, centralizes $(1\ 2)(3\ 4)$; there are $(n-4)!$ such elements. Similarly, all elements of the form $(1\ 4)(2\ 3)\tau$ centralize $(1\ 2)(3\ 4)$, for another $(n-4)!$, as do $(1\ 3\ 2\ 4)\tau$ and $(1\ 4\ 2\ 3)\tau$. These elements are all distinct, and total $8 \cdot (n-4)!$. Thus we have exhausted the centralizer of $(1\ 2)(3\ 4)$.

You counted $\tau$ three times. The two other permutation is that $(1\ 3\ 2\ 4)$ and $(1\ 4\ 2\ 3)$.