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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,

$\sigma_1 \sigma_2(a_1)= b_2$

$\sigma_1 \sigma_2(a_2)= b_1$

$\sigma_2 \sigma_1(a_1)= b_2$

$\sigma_2 \sigma_1(a_2)= b_1$

$\sigma_1 \sigma_2(b_1)= a_2$

$\sigma_1 \sigma_2(b_2)= a_1$

$\sigma_2 \sigma_1(b_1)= a_2$

$\sigma_2 \sigma_1(b_2)= a_1$

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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 $\sigma_i$.

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  • $\begingroup$ can you give me a numeral example so I can plug in values and see what's going on? I would really appreciate that! $\endgroup$ – combo student Oct 2 '16 at 22:39
  • $\begingroup$ Take an example and apply the proof it is constructive. $\endgroup$ – Tsemo Aristide Oct 2 '16 at 22:40
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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 ${\rm ord}(\sigma_s)=2$.

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