Characterize the conjugacy classes of Alt(n)

Let K \subseteq S_n be a conjugacy class and suppose that K \subseteq A_n.

  1. Show that \sigma \in S_n does not commute with any odd permutation if and only if the cycle type of \sigma consists of distinct odd integers.
  2. Deduce that K is a union of two A_n-conjugacy classes if and only if the cycle type of K consists of distinct odd integers.

[Hint: Assume first that \sigma \in K does not commute with any odd permutation. Observe that \sigma commutes with each cycle in its own cycle decomposition, so that each cycle must have odd length. If two cycles have the same odd length k, find a product of k transpositions which interchanges them and commutes with \sigma. Conversely, if the cycle type of \sigma consists of distinct integers, prove that \sigma commutes only with the subgroup generated by the cycles in its cycle decomposition.]


We begin with a remark.

Remark: If \sigma \in S_n is a cycle, then C_{S_n}(\sigma) = \langle \sigma, S_{F(\sigma)}, where F(\sigma) is the set of fixed points of \sigma and S_{F(\sigma)} is naturally identified with a subgroup of S_n.

Now to the main result.

(\Rightarrow) Suppose \sigma \in S_n does not commute with any odd permutation. Write \sigma = \alpha \tau, where \alpha is some cycle in the decomposition of \sigma and \alpha commutes with \tau. (Say \tau is the rest of the cycle decomposition of \sigma.) Note that \alpha \sigma = \alpha \alpha \tau = \alpha \tau \alpha = \sigma \alpha, so that \alpha commutes with \sigma. By our hypothesis, then, \alpha is not an odd permutation. Hence \alpha has odd length. Since \alpha was an arbitrary cycle in the decomposition of \sigma, the cycle type of \sigma consists of odd integers.

Now suppose some cycles \alpha and \beta in the decomposition of \sigma have the same length k, and write \sigma = \alpha \beta \theta for some \theta which commutes with both \alpha and \beta. (Again, say, the rest of the cycle decomposition of \sigma.) Now \alpha and \beta are conjugate in S_n, and moreover are disjoint. Write \alpha = (a_1\ a_2\ \ldots\ a_k) and \beta = (b_1\ b_2\ \ldots\ b_k); that is, define a and b such that \alpha(a_i) = a_{i+1} and \beta(b_i) = b_{i+1} for all 1 \leq i \leq k, where indices are taken mod k. Let \tau_i = (a_i\ b_{i+1}) for each 1 \leq i \leq k, and let \tau = \tau_1 \tau_2 \cdots \tau_k; note that \tau commutes with \theta. Moreover, by a previous exercise, we have M(\tau \alpha \tau^{-1}) = \tau[M(\alpha)] = M(\beta) by our definition of \tau. Now for each i, we have (\tau \alpha \tau^{-1})(b_i) = (\tau \alpha)(a_{i-1}) = \tau(a_i) = b_{i+1} = \beta(b_i), with indices taken mod k. Hence \tau \alpha \tau^{-1} = \beta. Moreover, since \tau is a product of disjoint 2-cycles, \tau^{-1} = \tau. Finally, note that \tau is an odd permutation, being an odd product of commuting transpositions. Now \tau \sigma = \tau \alpha \beta \theta = \beta \tau \beta \theta = \beta \alpha \tau \theta = \alpha \beta \theta \tau = \sigma \tau, so that \tau commutes with \sigma, a contradiction. So no two cycles in the decomposition of \sigma have the same length.

Thus the cycle type of \sigma consists of distinct odd integers.

(\Leftarrow) Suppose that the cycle type of \sigma consists of distinct odd integers. Then \sigma fixes at most one element; that is, |F(\sigma)| \leq 1. By the remark above, C_{S_n}(\sigma) = \langle \sigma, S_{F(\sigma)} \rangle. Now S_{F(\sigma)} consists of only the identity; hence C_{S_n}(\sigma) = \langle \sigma \rangle. Because \sigma is a product of even permutations, C_{S_n}(\sigma) consists only of even permutations. In particular, \sigma does not commute with any odd permutation.

Now for the second point.

Let K \subseteq A_n be a conjugacy class and let \sigma \in K.

Note that K consists of two A_n-conjugacy classes if and only if \sigma does not commute with any odd permutation by §4.3 #20, if and only if the cycle type of \sigma consists of distinct odd integers by the above argument.

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Comments

  • Gobi Ree  On December 12, 2011 at 12:33 am

    In (\Rightarrow), how about defining \tau=(a_1\ b_1) \cdots (a_k\ b_k)? Then the conjugation of \tau changes \alpha with \beta so that \tau \sigma \tau^{-1}=\sigma.

    • nbloomf  On December 15, 2011 at 11:08 am

      I like it.

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