Prove that $\sigma^2$=$\epsilon$ if and only if $\sigma$ can be expressed as a product of disjoint 2-cycles. $\epsilon$ denotes the identity permutation.
Any hints to help me get started would be really appreciated!
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Consider that for a permutation $\sigma$, $\sigma(\sigma^{-1}(x_n))=x_n$ implies, since also $\sigma(\sigma(x_n))=x_n$, implies that $\sigma=\sigma^{-1}$. Can you understand then what that means for the elements of the set being permuted? – Eleven-Eleven Feb 12 '16 at 02:11
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Hint. I assume you already know that any permutation is a product of disjoint cycles. Now suppose that one of these cycles is $$(\,a_1\ a_2\ \cdots\ a_n\,)$$ with $n\ge3$. Then $$\sigma^2(a_1)=\sigma(a_2)=a_3\ne a_1$$ and so $\sigma^2$ is not the identity.
David
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Consider a permutation $\sigma$ and elements $x_i, x_j, x_k$. If you assume that $\sigma^2=\epsilon$, then that means that $$\sigma^2(x_i)=x_i$$
Now consider that $\sigma$ sends $x_i$ to $x_j$ and sends $x_j$ to $x_k$. Thus
$$x_i=\sigma^2(x_i)=\sigma(\sigma(x_i))=\sigma(x_j)=x_k$$
Therefore we can represent this permutation as $(x_i\ x_j\,)$. As every other element $x_m, x_n$ have these properties, we have disjoint 2-cycles or we have 1-cycles for the condition that $\sigma_m=\sigma_n$.
Eleven-Eleven
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