Dtyjlui

Let be $(G,circ )$ a finite group and

$ pi : G rightarrow U(H) $ a group homomorphism .

Set

$V:= { xin H: pi(g)x=x, forall g in G } $

Then the orthogonal projection on $V$ is given as :

$$ frac{1}{|G|} sum_{ g in G} pi(g) $$

How can I verify that? Or how can start?
any help very appreciated ,as I am very stuck

Set

$P = displaystyle dfrac{1}{vert G vert} sum_{g in G} pi(g); tag 1$

for

$y in H, tag 2$

we have

$P^2y = displaystyle dfrac{1}{vert G vert} sum_{h in G} pi(h) left ( displaystyle dfrac{1}{vert G vert} sum_{g in G} pi(g) y right )$
$= dfrac{1}{vert G vert^2} displaystyle sum_{h in G} sum_{g in G} pi(h) pi(g)y = displaystyle dfrac{1}{vert G vert^2} sum_{h in G} sum_{g in G} pi(hg)y; tag 3$

now for any fixed $h$ in the rightmost sum of (3),

$displaystyle sum_{g in G} pi(hg)y = sum_{g in G} pi(g)y, tag 4$

since left multiplication by $h$ merely permutes the elements of $G$ amongst themselves; it follows that this rightmost sum of (3) is comprised of $vert G vert$ copies of $sum_{g in G} pi(g)y$; hence

$P^2y = dfrac{1}{vert G vert^2} vert G vert displaystyle sum_{g in G} pi(g)y = dfrac{1}{vert G vert} displaystyle sum_{g in G} pi(g)y = Py tag 5$

which shows that $P$ is a projection. $P$ is orthogonal if

$P^dagger = P; tag 6$

we have

$P^dagger = dfrac{1}{vert G vert} displaystyle sum_{g in G} pi(g)^dagger; tag 7$

since

$pi(g) in U(H), tag 8$

that is, $pi(g)$ is unitary,

$pi(g)^dagger = pi(g)^{-1} = pi(g^{-1}); tag 9$

thus

$P^dagger = dfrac{1}{vert G vert} displaystyle sum_{g in G} pi(g^{-1}) = dfrac{1}{vert G vert} sum_{g in G} pi(g) = P, tag{10}$

and $P$ is an orthogonal projection.