Help deriving that $\mathrm{sign} : S_n\to \{\pm 1\}$ is multiplicative

Question

$\def\sign{\operatorname{sign}}$

For homework, I am trying to show that the map $\sign:S_n \to \{\pm 1\}$ is multiplicative, i.e. that for any permutations $\sigma_1,\sigma_2$ in the symmetric group $S_n$, we have $$\sign(\sigma_1 \sigma_2) = \sign\sigma_1 \sign\sigma_2.$$

The definition for $\sign$ that I am using is that if $\sigma = \gamma_1\cdots \gamma_k$ is the cycle decomposition of $\sigma \in S_n$ and $\ell_1,\ldots,\ell_k$ are the cycle lengths of $\gamma_1,\ldots,\gamma_k$ respectively, then

$$\sign(\sigma):= (-1)^{\ell_1-1}\cdots(-1)^{\ell_k-1}.$$

I showed first that the formula holds for two transpositions. Then I showed that it holds for a transposition and a cycle.

However, I got stuck trying to show that the formula holds for a transposition and a product of two cycles, i.e. $\sigma_1 = \tau$ and $\sigma_2 = \gamma_1\gamma_2$. I feel like this case is much more complicated than the others which makes me think I am taking the wrong approach.

If $\tau,\gamma_1,\gamma_2$ are all disjoint then the formula holds trivially because then $\gamma_1\gamma_2\tau$ is the cycle decomposition of $\sigma_1\sigma_2$. Otherwise they are not all disjoint, at which point it seems to get complicated very quickly and I don't know how to proceed.

Would someone please help me understand the best approach to this proof?

Answer 1

It's not really that difficult to show it for a transposition that straddles two disjoint cycles. The key is to see that if we have an $n$-element cycle disjoint from a $k$-element cycle and compose with a transposition of one element in each cycle, then you always get an $(n+k)$-element cycle -- there's essentially only one way the transposition can connect the cycles. And then $(-1)^{n+k-1}=-1\times (-1)^{n-1}(-1)^{k-1}$.

Together with showing it for a transposition and a cycle that contains both transposed elements (which I imagine is exactly the reverse of the above case), this is all you need to show (by induction) that if a permutation is a product of $n$ transpositions, then its sign is $(-1)^n$.

For a general product of permutations, just decompose each factor into transpositions (you know you can always to this, right?) and count transpositions in each of them.

Answer 2

Hennings answer is a great way to go. I thought you might find this alternative viewpoint of interest as well...

Each permutation can be realized by a permutation matrix (let $\sigma \in S_n$ and take the $n\times n$ identity and then send row $i$ to row $\sigma(i)$). It's not hard to show that if $A$ and $B$ are matrices corresponding to $\sigma$ and $\tau$, then $AB$ corresponds to $\sigma \tau$.

Now recall that swapping rows changes the sign of the determinant. From here it's not hard to show that the sign of a permutation is just the determinant of the corresponding matrix.

Now the homomorphism property of $\mathrm{sign}$ comes from: $\mathrm{det}(AB)=\mathrm{det}(A)\mathrm{det}(B)$ (the fact that the determinant is a homomorphism).