The sign function is a homomorphism We define an inversion of a permutation $\sigma\in S_k$ to be a pair $(\sigma(i), \sigma(j))$ such that $i<j$ but $\sigma(i)> \sigma(j)$. The sign of $\sigma$, written $\text{sgn}(\sigma)$, is defined by 
\begin{align*}
\text{sgn}(\sigma) = (-1)^{\# \text{ of inversions in }\sigma}  = 
\begin{cases}
+1 &\text{ if the number of inversions in $\sigma$ is even}\\
-1 &\text{ if the number of inversions in $\sigma$ is odd}
\end{cases}.
\end{align*}
I want to prove that: $\text{sgn}(\sigma \tau)= (\text{sgn }\sigma)(\text{sgn }\tau)$ for any two permutations $\sigma$ and $\tau$, using the above definition. 
I tired many times but i failed. If I got some equation relating the number of inversions of $\sigma$, $\tau$ and the composite $\sigma\tau$, I had done. I need your help please.
 A: Yes, one can prove it directly by counting inversions. Starting with $i<j$, apply $\tau$ and get the pair $(i_1,j_1)$ defined by $$i_1=\tau(i)\qquad j_1=\tau(j).$$ Then apply $\sigma$ and get the pair $(i_2,j_2)$ defined by
$$i_2=\sigma(i_1)=\sigma\tau(i)\qquad j_2=\sigma(j_1)=\sigma\tau(j).$$
In summary
$$(i,j)\to^{\tau}(i_1,j_1)\to^\sigma(i_2,j_2) $$
After applying each permutation, an inversion either occurs or not.
Let $x$ count the number of pairs $i<j$ such that $i_1>j_1$ and $i_2<j_2$.
Let $y$ count the number of pairs $i<j$ such that $i_1>j_1$ and $i_2>j_2$.
Let $z$ count the number of pairs $i<j$ such that $i_1<j_1$ and $i_2>j_2$.
The permutations of interest then have the following numbers of inversions:
$$N(\tau)=x+y$$
$$N(\sigma)=x+z$$
$$N(\sigma\tau)=y+z$$
It follows that
$$sgn(\sigma)sgn(\tau)=(-1)^{N(\sigma)}(-1)^{N(\tau)}=(-1)^{x+z}(-1)^{x+y}$$
$$=(-1)^{2x+y+z}=(-1)^{y+z}=(-1)^{N(\sigma\tau)}=sgn(\sigma\tau)$$
A: Another way to do this is to use the following argument, which I believe is the one Foote uses in his abstract algebra book:
define $\Delta = \prod_{1\le i<j\le n} (x_i-x_j)$, and put $\sigma(\Delta) = \prod_{1\le i<j\le n} (x_{\sigma(i)}-x_{\sigma(j)})$. Then, it's not too hard to show,  that $\text{sgn}(\sigma)=\frac{\Delta}{\sigma(\Delta)}$ and that $\text{sgn}:S_n\to \left \{ -1,1 \right \}$ is a homomorphism of groups. (One starts by noting the effect of a given transposition on $\Delta$).
A: Let $N(\rho)$ be the number of inversions in $\rho$ for any permutation $\rho$. First, note that any permutation $\rho$ can be written as $N(\rho)$ transpositions.
This means $\sigma$ can be written with $N(\sigma)$ transpositions and $\tau$ can be written with $N(\tau)$ transpositions. Then, when we put the two together to get $\sigma\tau$, we are writing it by putting all of the transpositions together, meaning we just wrote $\sigma\tau$ with $N(\sigma)+N(\tau)$ transpositions. However, we can also write $\sigma\tau$ with $N(\sigma\tau)$ transpositions as according to our note at the beginning.
Now, there is a theorem that tells us that the numbers of inversions a certain permutation can be written as is either all even or all odd, and thus they are all $\equiv \pmod 2$. Since $\sigma\tau$ can be written both as $N(\sigma)+N(\tau)$ and $N(\sigma\tau)$ transpositions, we have:
$$N(\sigma\tau) \equiv N(\sigma)+N(\tau) \pmod 2$$
Thus, we have the following:
$$\text{sgn}(\sigma\tau)=(-1)^{N(\sigma\tau)}$$
Because of $N(\sigma\tau) \equiv N(\sigma)+N(\tau) \pmod 2$ and the fact that $a \equiv b \pmod 2$ implies $(-1)^a=(-1)^b$:
$$=(-1)^{N(\sigma)+N(\tau)} \\=(-1)^{N(\sigma)}(-1)^{N(\tau)} \\=\text{sgn}(\sigma)\text{sgn}(\tau)$$
A: Let $S_n$ act on the set $X = \{1, \dots , n\}.$ For $g\in S_n$, define $\operatorname{sgn}(g) = (-1)^{n+r(g)}$, where $r(g) = \#X/\langle{g\rangle}$ denotes the number of distinct orbits of $X$ under the action of $g$. Now check that for any transposition $\sigma = (ab)\in S_n$, we have
\begin{align*}
r(g\sigma) = \begin{cases}
r(g) + 1 & \text{$a, b$ belong to the same orbit of $S_n$ under $\langle{g\rangle}$};  \\
r(g) - 1 & \text{$a, b$ belong to distinct orbits of $S_n$ under $\langle{g\rangle}$}.\\
\end{cases}
\end{align*}
A: One can define $\epsilon_n: S_n \rightarrow \{1,-1\}$ inductively by letting $\varepsilon_1$ be the trivial homomorphism and for $n \geq 1$
$$\varepsilon_{n+1}(\sigma)= \begin{cases} \varepsilon_n(\sigma) \text{ if } \sigma \in S_n \\
-\varepsilon_n((n+1 \text{ } \sigma(n+1)) \sigma)  \text{ if } \sigma \notin S_n \end{cases}.$$
One now proves, with some case distinction and induction, that we have, for all $1 \leq i < j \leq n$,
$$\varepsilon_n((i \text{ } j) \sigma)=-\varepsilon_n(\sigma)$$
and 
$$\varepsilon_n(\text{id})=1$$
from which it will follow that $\varepsilon_n$ is indeed the sign function from your question, which now is automatically well-defined since $\varepsilon_n$ is well-defined.
