In $S_n$, if $ε = α_1α_2 \cdots α_r$ where $α_i$ is a $2$-cycle, then $r$ is even.

Question

In $S_n$, if $ε = α_1α_2 \cdots α_r$ where $α_i$ is a $2$-cycle, then $r$ is even.

I don't know how to start. Note, $ε$ is the identity of the permutation group $S_n$.

Answer 1

Any permutation in $S_n$ can be written as the product of $2$-cycles (often called transpositions), but there are many ways to do this. However, given $\sigma \in S_n$, it cannot be written as both the product of an even number of transpositions and the product of an odd number of transpositions. That is, if $\sigma = \alpha_1\alpha_2\dots\alpha_r$ and $\sigma = \beta_1\beta_2\dots\beta_s$ where $\alpha_1, \alpha_2, \dots, \alpha_r, \beta_1, \beta_2, \dots, \beta_s$ are transpositions, then $r$ and $s$ are both even or $r$ and $s$ are both odd. A proof of this fact can be found here. If $\sigma$ can only be written as the product of an even/odd number of transpositions, we say that $\sigma$ is an even/odd permutation. This is sometimes called the parity of the permutation $\sigma$.

Now consider the identity element of $S_n$ which you have denoted $\epsilon$. I claim that $\epsilon$ is an even permutation. To demonstrate this, I need to display one way of writing $\epsilon$ as the product of an even number of transpositions. Well, you could consider $\varepsilon$ as a product of zero transpositions, but if you're not comfortable with that, let $\alpha$ be any transposition, then $\epsilon = \alpha\alpha$. As there are two transpositions in the product, and two is even, $\epsilon$ is an even permutation. Therefore, if $\epsilon = \alpha_1\alpha_2\dots\alpha_r$ where $\alpha_1, \alpha_2, \dots, \alpha_r$ are transpositions, then $r$ must be even.