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


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:


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.