If $\sigma$ is an odd permutation, explain why $\sigma^2$ is even but $\sigma^{-1}$ is odd.

Question

If $\sigma$ is an odd permutation, why is $\sigma^2$ even and $\sigma^{-1}$ odd?

Edit: If $\sigma$ is even, then ${\rm sign}(\sigma)= 1$, and if $\sigma$ is odd, then ${\rm sign}(\sigma) = -1$. Based on this definition, would the answer be that $(-1)^2 = 1$ thus even, and $(-1)^{-1} = \frac{1}{-1} = -1$ thus odd?

Answer 1

I'm a little rusty at this stuff, but as I recall:

1.) Permutations may be represented by permutation matrices, which are $0$-$1$ matrices having exactly one $1$ in each row and column;

2.) $P$ is the matrix of an even permutation if and only if

$\det P = 1; \tag 1$

3.) $P$ is the matrix of and odd permutation if and only if

$\det P = -1; \tag 2$

4.) Composition of permutations corresponds to multiplication of their matrices; indeed, there exist isomorphisms

$\phi:S_n \to GL_n(\Bbb Z) \tag 3$

taking permutations to their corresponding matrices; $\phi(S_n)$ is a subgroup of $GL_n(\Bbb Z)$; such isomorphisms are not unique since a permutaton of an ordered $n$-tuple such as $(1, 2, \ldots, n)$ on which the action of $S_n$ is defined may give rise to different isomorphisms.

I believe these assertions are quite well known and may be found in many books on abstract algebra and/or group theory.

We denote the identity permutation, which fixes everything, by $P_{Id}$.

If we accept the above, and if $\sigma$ is an odd permutation, then a matrix $P_\sigma$ representing $\sigma$ satisfies

$\det P_\sigma = -1; \tag 4$

thus

$\det P_{\sigma^2} = (\det P_\sigma)^2 = (-1)^2 = 1, \tag 5$

so $\sigma^2$ is even; also

$(\det P_\sigma)(\det P_{\sigma^{-1}}) = \det(P_\sigma P_{\sigma^{-1}})$ $= \det P_{\sigma \sigma^{-1}} = \det P_{Id} = \det I = 1, \tag 6$

and with odd $\sigma$,

$\det P_\sigma = -1; \tag 7$

this yields

$\det P_{\sigma^{-1}} = -1, \tag 8$

so $\sigma^{-1}$ is also odd.

Perhaps not the most elegant or rigorous explanation, but it may be adequate for the present purposes.

Answer 2

Since $\sigma$ is an odd permutation, in particular, it can be written as a product of an odd number of transpositions. (This is an equivalent way to define an odd permutation. Why?) Thus $\sigma^2$ can be written as a product of an even number of transpositions, so $\sigma^2$ is an even permutation. (Why?) The reason why $\sigma^{-1}$ is odd is that the inverse of a transposition is itself.

Answer 3

The map sign that assigns to every permutation its signature is a group homomorphism from the symmetric group to the multiplicative group $\{−1, 1\}$: $$ \text{sign}: S_n \to \{−1, 1\}. $$ By definition, a permutation is even if its signature is $1$ and a permutation is is odd if its signature is $-1$. Therefore, if $\sigma$ is an odd permutation, $$ \text{sign}(\sigma^2) = \text{sign}(\sigma)^2 = (-1)^2 = 1 \quad\text{and}\quad \text{sign}(\sigma^{-1}) = \text{sign}(\sigma)^{-1} = (-1)^{-1} = -1 $$ Thus $\sigma^2$ is even and $\sigma^{-1}$ is odd.