# Proving that a permutation $\sigma \in S_n$ of order two is a product of disjoint 2-cycles

Let $\sigma \in S_n$ be a permutation of order two. Prove that $\sigma$ is a product of disjoint 2-cycles.

My attempt:

If $\sigma \in S_n$ is a permutation of order two then $S_n$ has cycles of at most length $k=2$. By definition, two cycles are called disjoint if $\alpha = (a_1,a_2,...a_l)$ and $\theta = (b_1,b_2,...b_k)$ such that $a_i \neq b_j$, so for this case,

Let

$\sigma_1 = (a_1,a_2)$ and $\sigma_2= (b_1,b_2)$

$a_1\neq b_1$ and $a_2 \neq b_2$

Then,

$\sigma_1 \sigma_2(a_1)= b_2$

$\sigma_1 \sigma_2(a_2)= b_1$

$\sigma_2 \sigma_1(a_1)= b_2$

$\sigma_2 \sigma_1(a_2)= b_1$

$\sigma_1 \sigma_2(b_1)= a_2$

$\sigma_1 \sigma_2(b_2)= a_1$

$\sigma_2 \sigma_1(b_1)= a_2$

$\sigma_2 \sigma_1(b_2)= a_1$

Suppose $\sigma$ is a permutation of $\{1,...,n\}$, you have a partition of $\{1,..,n\}$ with $\{a_i,b_i\}$ where $\sigma(a_i)=b_i$. Let $\sigma_i$ the cycle $(a_i,b_i)$ such that $a_i\neq b_i$, $\sigma$ is the product of the $\sigma_i$.
Every permutation has a decomposition into disjoint cycles. Remember that if $$\sigma=\sigma_1\sigma_2...\sigma_k$$, with $$\sigma_i,\sigma_j$$ disjoint if $$i\neq j$$, then $$\sigma$$'s order is equal to $${\rm lcm}({\rm ord}(\sigma_1),{\rm ord}(σ_2),\dots,{\rm ord}(\sigma_k))$$. If $${\rm ord}(\sigma)=2$$ then $$\exists\sigma_s, ~s=1,\dots,k$$, such that $${\rm ord}(\sigma_s)=2$$.