Characterization of elements of order 2 in Sym(n)

Show that an element has order 2 in S_n if and only if its cycle decomposition is a product of commuting 2-cycles.


Lemma 1. Let G be a group and I a finite nonempty index set. For all n \in \mathbb{Z} and \{a_i\}_I \subseteq G mutually commute we have ( \prod_I a_i )^n = \prod_I a_i^n. Proof: We proceed by induction on the cardinality of I. If |I| = 1 the statement is trivial. If the statement holds for all indexing sets of cardinality n, and I is an index set of cardinality n+1, we have (\prod_I a_i)^n = (a_j \cdot \prod_{I \setminus \{j\}} a_i)^n = a_j^n \cdot (\prod_{I \setminus \{j\}} a_i)^n = a_j^n \cdot \prod_{I \setminus \{j\}} a_j^n = \prod_I a_i^n. Thus by induction the statement holds for all finite sets I. \square

Now to the main result.

(\Leftarrow) Suppose \sigma \in S_n is a (necessarily finite) product \sigma = \prod \sigma_i of commuting 2-cycles. Then by the lemma, \sigma^2 = (\prod \sigma_i)^2 = \prod \sigma_i^2 = \prod 1 = 1. Thus |\sigma| = 2.

(\Rightarrow) Let \sigma \in S_n be an element of order 2, and suppose that the cycle decomposition of \sigma is not a product of commuting 2-cycles. Then some cycle in the decomposition of \sigma has length at least 3; say this cycle is (a_0\ a_1\ a_2\ \ldots\ a_k). Note that \sigma^2(a_0) = a_2 \neq a_0, so that \sigma^2 \neq 1. This is a contradiction; in particular, every cycle in the decomposition of \sigma has length 2. So the cycle decomposition of \sigma is a product of commuting 2-cycles.

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Comments

  • Eric Ling  On October 25, 2011 at 11:38 pm

    Hi,

    For the forward direction, I believe you only proved that sigma must be the product of 2-cycles, but you haven’t showed the commutative part.

    • nbloomf  On October 26, 2011 at 2:47 pm

      What we showed here is that the cycle decomposition of \sigma consists of 2-cycles. The cycles in this decomposition commute since they are disjoint.

      I edited the proof to make this more clear. Thanks!

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