Permutation/Sign/Group homomorphism/Fact/Proof
< Permutation < Sign < Group homomorphism <
Fact
Proof
Let two permutations
π
{\displaystyle {}\pi }
and
ρ
{\displaystyle {}\rho }
be given. Then
sgn
(
π
∘
ρ
)
=
∏
i
<
j
(
π
∘
ρ
)
(
j
)
−
(
π
∘
ρ
)
(
i
)
j
−
i
=
(
∏
i
<
j
(
π
∘
ρ
)
(
j
)
−
(
π
∘
ρ
)
(
i
)
ρ
(
j
)
−
ρ
(
i
)
)
∏
i
<
j
ρ
(
j
)
−
ρ
(
i
)
j
−
i
=
(
∏
i
<
j
,
ρ
(
i
)
<
ρ
(
j
)
π
(
ρ
(
j
)
)
−
π
(
ρ
(
i
)
)
ρ
(
j
)
−
ρ
(
i
)
)
(
∏
i
<
j
,
ρ
(
i
)
>
ρ
(
j
)
π
(
ρ
(
j
)
)
−
π
(
ρ
(
i
)
)
ρ
(
j
)
−
ρ
(
i
)
)
sgn
(
ρ
)
=
(
∏
i
<
j
,
ρ
(
i
)
<
ρ
(
j
)
π
(
ρ
(
j
)
)
−
π
(
ρ
(
i
)
)
ρ
(
j
)
−
ρ
(
i
)
)
(
∏
i
<
j
,
ρ
(
i
)
>
ρ
(
j
)
π
(
ρ
(
i
)
)
−
π
(
ρ
(
j
)
)
ρ
(
i
)
−
ρ
(
j
)
)
sgn
(
ρ
)
=
∏
k
<
ℓ
π
(
ℓ
)
−
π
(
k
)
ℓ
−
k
sgn
(
ρ
)
=
sgn
(
π
)
sgn
(
ρ
)
.
{\displaystyle {}{\begin{aligned}\operatorname {sgn} (\pi \circ \rho )&=\prod _{i<j}{\frac {(\pi \circ \rho )(j)-(\pi \circ \rho )(i)}{j-i}}\\&={\left(\prod _{i<j}{\frac {(\pi \circ \rho )(j)-(\pi \circ \rho )(i)}{\rho (j)-\rho (i)}}\right)}\prod _{i<j}{\frac {\rho (j)-\rho (i)}{j-i}}\\&={\left(\prod _{i<j,\,\rho (i)<\rho (j)}{\frac {\pi (\rho (j))-\pi (\rho (i))}{\rho (j)-\rho (i)}}\right)}{\left(\prod _{i<j,\,\rho (i)>\rho (j)}{\frac {\pi (\rho (j))-\pi (\rho (i))}{\rho (j)-\rho (i)}}\right)}\,\operatorname {sgn} (\rho )\\&={\left(\prod _{i<j,\,\rho (i)<\rho (j)}{\frac {\pi (\rho (j))-\pi (\rho (i))}{\rho (j)-\rho (i)}}\right)}{\left(\prod _{i<j,\,\rho (i)>\rho (j)}{\frac {\pi (\rho (i))-\pi (\rho (j))}{\rho (i)-\rho (j)}}\right)}\,\operatorname {sgn} (\rho )\\&=\prod _{k<\ell }{\frac {\pi (\ell )-\pi (k)}{\ell -k}}\operatorname {sgn} (\rho )\\&=\operatorname {sgn} (\pi )\operatorname {sgn} (\rho ).\end{aligned}}}
To fact
and 
be given. Then