I am trying to show that if $\sigma \in S_n$, $\sigma^2 = \epsilon$ iff $\sigma$ is product of disjoint transpositions.

This is my attempt:

Suppose $\sigma = (k_1\ k_2)\dots(k_{r-1}\ k_r)$ then the result follows from the fact that disjoint transpositions commute and and that $(k_i\ k_j)^2 = \epsilon$.

Conversely, suppose that $\sigma^2 = \epsilon$. Then $\sigma = \sigma^{-1}$. Write $\sigma = (k_1 \dots k_r)$, thus $(k_1 \dots k_r) = (k_r \dots k_1)$. From this we see that:

$k_1 \rightarrow k_2$ on the left and $k_2 \rightarrow k_1$ on the right. This realtion holds for all $k_i$ which implies the result.

Is this proof correct? I am unsure about the converse, but pretty sure the first part is solid. Thanks.

  • $\begingroup$ You assume that $\sigma$ is a cycle, why? $\endgroup$ – Najib Idrissi Sep 12 '15 at 8:37
  • $\begingroup$ Good point. I guess to fix this we can generalize the converse as any $\sigma$ can be written as the product of disjoint cycles of length at least 2 (Cycle Decomposition Theorem). $\endgroup$ – dable Sep 12 '15 at 8:45
  • $\begingroup$ If $\sigma=\sigma_1\cdots\sigma_k$ is a decomposition of $\sigma$ as a product of disjoint cycles of length $\ell_1,\ldots,\ell_k$, then $|\sigma|=\mathrm{lcm}(\ell_1,\ldots,\ell_k)$. The converse follows. $\endgroup$ – David Hill Sep 12 '15 at 14:19
  • $\begingroup$ How does the converse follow? $\endgroup$ – dable Sep 12 '15 at 14:43
  • $\begingroup$ Also, if by $|\sigma|$ you mean the number of elements in $\sigma$ then this does not hold for $\sigma = (1 \ 2)(3 \ 4 \ 5)$. $\endgroup$ – dable Sep 12 '15 at 14:49

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