Dtyjlui

We define an inversion of a permutation $sigmain S_k$ 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 $sigmatau$, I had done. I need your help please.

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
$$i_2=sigma(i_1)=sigmatau(i)qquad j_2=sigma(j_1)=sigmatau(j).$$
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:
$$N(tau)=x+y$$
$$N(sigma)=x+z$$
$$N(sigmatau)=y+z$$
It follows that
$$sgn(sigma)sgn(tau)=(-1)^{N(sigma)}(-1)^{N(tau)}=(-1)^{x+z}(-1)^{x+y}$$

$$=(-1)^{2x+y+z}=(-1)^{y+z}=(-1)^{N(sigmatau)}=sgn(sigmatau)$$

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 $Delta = prod_{1le i<jle n} (x_i-x_j)$, and put $sigma(Delta) = prod_{1le i<jle n} (x_{sigma(i)}-x_{sigma(j)})$. Then, it's not too hard to show, that $text{sgn}(sigma)=frac{Delta}{sigma(Delta)}$ and that $text{sgn}:S_nto left { -1,1 right }$ is a homomorphism of groups. (One starts by noting the effect of a given transposition on $Delta$).

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 $sigmatau$, we are writing it by putting all of the transpositions together, meaning we just wrote $sigmatau$ with $N(sigma)+N(tau)$ transpositions. However, we can also write $sigmatau$ with $N(sigmatau)$ 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 $sigmatau$ can be written both as $N(sigma)+N(tau)$ and $N(sigmatau)$ transpositions, we have:
$$N(sigmatau) equiv N(sigma)+N(tau) pmod 2$$

Thus, we have the following:
$$text{sgn}(sigmatau)=(-1)^{N(sigmatau)}$$
Because of $N(sigmatau) equiv N(sigma)+N(tau) pmod 2$ and the fact that $a equiv b pmod 2$ implies $(-1)^a=(-1)^b$:
$$=(-1)^{N(sigma)+N(tau)} \=(-1)^{N(sigma)}(-1)^{N(tau)} \=text{sgn}(sigma)text{sgn}(tau)$$

Let $S_n$ act on the set $X = {1, dots , n}.$ For $gin S_n$, define $operatorname{sgn}(g) = (-1)^{n+r(g)}$, where $r(g) = #X/langle{grangle}$ 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(gsigma) = begin{cases}
r(g) + 1 & text{$a, b$ belong to the same orbit of $S_n$ under $langle{grangle}$}; \
r(g) - 1 & text{$a, b$ belong to distinct orbits of $S_n$ under $langle{grangle}$}.\
end{cases}
end{align*}

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.