Prove this result relating to the sign of a permutation Suppose that $\phi \in S_n$ is a permutation.
Suppose also that $\psi = \phi \circ (i,j),$ where $1 \leq i, j \leq n.$
Why does it follow that sign$(\phi) = $ $-$sign$(\psi)$?
 A: With the definition of sign in terms of transpositions, this is a one line proof. Namely, if $\phi$ is a product of an odd number of transpositions, then $\psi$ is the product of an even number of transpositions, and vice versa. Hence they have opposite signs.
Less trivial is with the definition in terms of inversions (odd number of inversions means sign $-1$, even number sign $1$). Suppose $\phi$ has $n$ inversions. Define
$a$= the number of indices $k$ with $i<k<j$ such that $\phi(i)>\phi(k)$
$b$= the number of indices $k$ with $i<k<j$ such that $\phi(i)<\phi(k)$
$c$= the number of indices $k$ with $i<k<j$ such that $\phi(j)>\phi(k)$
$d$= the number of indices $k$ with $i<k<j$ such that $\phi(j)<\phi(k)$
Then the number of inversions of $\psi$ is 
$$n-a+b+c-d+1$$
But $a+b=c+d=j-i-1$, because they are both equal to the number of indices between $i$ and $j$, so $-a+b+c-d$, which has the same parity as $a+b+c+d$, is even, hence the number of inversions of $\psi$ has the opposite parity of the number of inversions of $\phi$ (from the $+1$ summand) hence they have opposite signs.
