Show that $\sigma$ has order 2 if and only if $\sigma$ is a product of disjoint 2-cycles.

Question

Here, $\sigma \in S_n$. The backwards (<=) direction is fairly straightforward. Assume $\sigma$ is a product of disjoint 2-cycles. Then the order of $\sigma$ is simply lcm$(2,2,2,...) = 2$. Conversely, if we assume $\sigma$ has order 2, I'm not sure how to show that it necessarily follows that $\sigma$ is a product of disjoint 2-cycles. However, I do think that it does have something to do with the fact that 2 is prime.

Answer 1

Hint: Every permutation can be written as a product of disjoint cycles. What happens if one of these cycles has length $>2$?

Answer 2

Suppose $\sigma$ has order 2. Then $\sigma^2$ is the identity permutation. Suppose to the contrary in the disjoint cycle representation there exists a cycle of length $k>2$, say $(a_1, a_2, \dots, a_k)$. Then $\sigma^2$ maps $a_1$ to $a_3 \neq a_1$, so $\sigma^2 \neq e$, where $e$ represents the identity permutation. Contradiction.