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

Lemma 1. Let be a group and a finite nonempty index set. For all and mutually commute we have . Proof: We proceed by induction on the cardinality of . If the statement is trivial. If the statement holds for all indexing sets of cardinality , and is an index set of cardinality , we have . Thus by induction the statement holds for all finite sets .

Now to the main result.

Suppose is a (necessarily finite) product of commuting 2-cycles. Then by the lemma, . Thus .

Let be an element of order 2, and suppose that the cycle decomposition of is not a product of commuting 2-cycles. Then some cycle in the decomposition of has length at least 3; say this cycle is . Note that , so that . This is a contradiction; in particular, every cycle in the decomposition of has length 2. So the cycle decomposition of is a product of commuting 2-cycles.

## Comments

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.

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

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