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Show that the sign representation of $S_n$ is indeed a representation.

attempt: Recall the sign function of a permutation is given by $\mathrm{sgn}(\pi) = (-1)^k$.

Then recall a representation is a homomorphism.

So we we have to show $\mathrm{sgn}(\pi_1 \pi_2)$ = $\mathrm{sgn}(\pi)\mathrm{sgn}(\pi_2)$.

This is a 3 part problem.

part a) If $\pi$ is a product of $k$ transpositions , then $k \equiv \mathrm{inv}(\pi)\pmod{2}$. Where $\mathrm{inv}(\pi)$ is the number of inversions of $\pi$.

part b) Conclude that the sign of a permutation is well defined.

So c) says "Conclude that the sign representation of $S_n$ is indeed a representation.

I am not sure what that representation might be. Or what I have to show. Can I use part $a )$ , or $b)$ to conclude $c$?

Is this what I have to show? Thanks for any feedback!

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    $\begingroup$ Well, what definition of representation have you been given? $\endgroup$ – Tobias Kildetoft Feb 9 '16 at 12:16
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We consider permutations $\pi,\sigma$ of $[n]:=\{1,2,\dots, n\}$. All sums below run over pairs $(i,j)$ in $[n]\times[n]$.

Using Iverson brackets, define the number of inversions of $\pi$ as $$\mbox{inv}(\pi)=\sum [i>j]\,[\pi(j)>\pi(i)].$$ We rewrite the sum above, then split it into two as below: \begin{eqnarray*} \mbox{inv}(\pi)&=&\sum [i>j]\,[\pi(j)>\pi(i)]\\ &=&\sum [\sigma(i)>\sigma(j)]\,[\pi\sigma(j)>\pi\sigma(i)]\\ &=&\sum [i>j]\, [\sigma(i)>\sigma(j)]\,[\pi\sigma(j)>\pi\sigma(i)]\\ &+&\sum [j>i]\,[\sigma(i)>\sigma(j)]\,[\pi\sigma(j)>\pi\sigma(i)]. \end{eqnarray*} Therefore (notice the switch of variables on the second sum) we have $$\mbox{inv}(\pi)=\sum [i>j]\, [\sigma(i)>\sigma(j)]\,[\pi\sigma(j)>\pi\sigma(i)]+\sum [i>j]\,[\sigma(j)>\sigma(i)]\,[\pi\sigma(i)>\pi\sigma(j)].$$ Similarly, splitting the sum defining $\mbox{inv}(\sigma)$ gives $$\mbox{inv}(\sigma)=\sum [i>j]\, [\sigma(j)>\sigma(i)]\,[\pi\sigma(j)>\pi\sigma(i)]+\sum [i>j]\,[\sigma(j)>\sigma(i)]\,[\pi\sigma(i)>\pi\sigma(j)].$$ Adding these two sums gives $$\mbox{inv}(\pi)+\mbox{inv}(\sigma)=\mbox{inv}(\pi\sigma)+ 2\sum [i>j]\,[\sigma(j)>\sigma(i)]\,[\pi\sigma(i)>\pi\sigma(j)].$$

Thus the map $\pi\mapsto \mbox{sign}(\pi):=(-1)^{\mbox{inv}(\pi)}$ is a homomorphism from $S_n$ to $\{\pm 1\}$.

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