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Prove for $\sigma \in S_n$ that $\sigma^2 = e$ if an only if $\sigma$ is a product of disjoint transpositions.

Workings:

I'm not too sure about this problem.

I know $e$ is the identity permutation which is the following:

$\begin{pmatrix} 1 & 2 & 3 & ... & n \\ 1 & 2 & 3 & ... & n \end{pmatrix}$

And $\sigma$ I think would look something like:

$(a_1,a_2)(a_3,a_4)...(a_{n-1},a_n)$

So for the if statement I would have to show that the above multiplied by itself gives $e$.

Though I am not sure how to do that. And I am unsure about the "only if" part.

Any help will be appreciated.

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  • $\begingroup$ By "trajectories" in the title did you mean "transpositions"? $\endgroup$ – Matt Samuel Mar 5 '15 at 19:29
  • $\begingroup$ Yes, sorry about that. $\endgroup$ – CoolNewFriends Mar 5 '15 at 19:30
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Any permutation is a product of disjoint cycles. Say $\sigma=C_1C_2\cdots C_k$ is a decomposition into such a product. Then $$\sigma^2=C_1^2C_2^2\cdots C_k^2$$ since the cycles commute with each other. If $C_i=(c_1c_2\cdots c_{2p})$ is a cycle of even length, then $$C_i^2=(c_1c_3c_5\cdots c_{2p-1})(c_2c_4c_6\cdots c_{2p})$$ If on the other hand $C_i=(c_1c_2\cdots c_{2p-1})$ is a cycle of odd length, then $$C_i^2=(c_1c_3c_5\cdots c_{2p-1}c_2c_4c_6\cdots c_{2p-2})$$ Thus $$\sigma^2=C_1'C_1''C_2'C_2''\cdots C_k'C_k''$$ where $C_i'$ and $C_i''$ are the two disjoint cycles in $C_i^2$ if $C_i$ is an even cycle, and $C_i'=C_i^2$ and $C_i''=e$ if $C_i$ is an odd cycle. This is a product of disjoint cycles, hence there are no cancellations. Thus the whole product is the identity if and only if each $C_i$ squares to the identity, meaning that $\sigma$ is a product of disjoint transpositions, which is what we wanted to show.

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  • $\begingroup$ Okay thanks for the help. $\endgroup$ – CoolNewFriends Mar 5 '15 at 19:34
  • $\begingroup$ @CoolNewFriends no problem. $\endgroup$ – Matt Samuel Mar 5 '15 at 19:36

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