## 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!