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).

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Comments

  • Gobi Ree  On December 10, 2011 at 10:31 am

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

    • nbloomf  On December 12, 2011 at 11:37 am

      Thanks!

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