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

Solution 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.

Solution 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).