Prove that $sgn(\sigma_1 \circ \sigma_2) = sgn(\sigma_1)sgn(\sigma_2)$

Lete $n\in \mathbb{N}$. Show that the transformation $$sgn: S_n \rightarrow \{\pm 1\}$$

(where $S_n$ is the set of all permutations of the integers in the set $\{1,...,n\}$),given by $\sigma \mapsto sgn(\sigma)$, $\sigma \in S_n$, is a group homomorphism, i.e. the following equality holds:

$$sgn(\sigma_1 \circ \sigma_2) = sgn(\sigma_1)sgn(\sigma_2)$$

Def.(sgn) : Let $\sigma \in S_n$ be a permutation that can be expressed as l transpositions. Then $sgn(\sigma) := (-1)^l$

My attempt:

Since every permutation $\sigma$ can be expressed as the product of a given number of transpositions, I have the following:

Let $\sigma_1 = \tau_1 \circ \tau_2 \circ .... \circ \tau_l$ and $\sigma_2 = \tau_1 \circ \tau_2 \circ .... \circ \tau_k$. Then it follows that $\sigma_1 \circ \sigma_2 = \tau_1 \circ \tau_2 \circ .... \circ \tau_l \circ \tau_1 \circ \tau_2 \circ .... \circ \tau_k$. Now I'm not exactly sure how to go from here and whether my approach is correct. I kind of doubt it, since I need $l + k$ transpositions to derive $\sigma_1 \circ \sigma_2$, but the sum of those could easily be greater than $n$, right? In case my approach was correct thus far, how can I show that the equality holds, could I case cases depending on whether l and k are even or odd to prove the equality?

• How is $sgn$ defined? As long as that is unclear it is impossible to prove that it is a grouphomomorphism. – H.B. Apr 17 '14 at 17:38
• I edited the post. – eager2learn Apr 17 '14 at 17:45

The hard part of $sgn$ is to show that any transposition decomposition of a permutation will always have an odd number or and even number of transposition, regardless of how you decompose it--you can have as many transpositions as you want, even more than $n$ transpositions. So you assign permutations with an even number of transpositions 1 and permutations with an odd number of transpositions the number -1.