# Math Challenge - May 2021

• Challenge
Sure, but what if ##z\neq 0##?
then,
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
\begin{align} 2x=z(a+b)\nonumber\\ 2y=z(a-b)\nonumber \end{align}
and putting ##2cz=(a^2+b^2)z^2## ⇒ ##z=\frac {2c} {a^2+b^2}##in these two equations, we get the above values of (x,y,z)

Problem #12
$$y^2=x\cdot (x+1)\cdot (x+7)\cdot (x+8)$$
substitute ##x+4 \rightarrow t##
then the equation becomes,
\begin{align} y^2&=(t-4)\cdot(t-3)\cdot(t+3)\cdot(t+4)\nonumber\\ y^2&=(t^2-16)\cdot(t^2-9)\nonumber \end{align}
so, for the R.H.S to be a perfect square,
the only possibilities are ##t=3,4,5##

as for product of two numbers (say ##a,b##) to be a perfect square, the only possibilities are,
if ##a=b,a=0,b=0\space or\space a=l^2,b=m^2## (where l,m are any real numbers)

if we use ##t^2-16=t^2-9##, then we won't get any solutions, so we'll have to use
##t^2-16=l^2\space \text{&} \space t^2-9=m^2##
from this we get,
##t^2=16+l^2=m^2+9##
clearly ##t=5## is the only possibility as 3,4,5 are pythagorean triplets.

Also we can have either ##t^2-9=0## or ##t^2-16=0##
from this we get  ##t=3,4,-3,-4## 

So putting the obtained values back in ##t=x+4## we get  ##x=-8,-7,-1,0,1## 

So ordered pairs ##(x,y)## are  ##(-8,0);(-7,0);(-1,0);(0,0);(1,144);(1,-144)## 

*Edited the answer to include all values of (x,y)

Last edited:
Mentor
then,
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
\begin{align} 2x=z(a+b)\nonumber\\ 2y=z(a-b)\nonumber \end{align}
and putting ##2cz=(a^2+b^2)z^2## ⇒ ##z=\frac {2c} {a^2+b^2}##in these two equations, we get the above values of (x,y,z)
This works only for ##a^2+b^2\neq 0##. Your first post was already correct, except for one special case.

This works only for ab≠0.
Why? What is the problem if ab=0, we should still have
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$

Edit: I see that if both a & b are 0 then this is wrong. So yes both a & b shouldn't be zero , but one of them can be right?

Is a=b=0 the special case you were talking about here?

Mentor
Is a=b=0 the special case you were talking about here?
Yes. ##a=b=c=0## and ##x=y=0## is a possibility, where ##z## doesn't have to be zero.

Yes. ##a=b=c=0## and ##x=y=0## is a possibility, where ##z## doesn't have to be zero.
But if c=0 then z is also 0

Mentor
But if c=0 then z is also 0
No. If ##a=b=c=0## and ##x=y=0## then ##z=1## is a solution, as is any arbitrary value for ##z##.

kshitij
No. If ##a=b=c=0## and ##x=y=0## then ##z=1## is a solution, as is any arbitrary value for ##z##.
Yes I missed that, I was looking at this expression
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
Didn't even notice the question, my bad.

Sorry for the dumb question but do the ##\circ##'s in 4) mean function composition or matrix multiplication? As I understand it, ##\tau(g)## and ##\rho(g^{-1})## are matrices in ##GL(W)## and ##GL(V)## resp. ?

Mentor
Sorry for the dumb question but do the ##\circ##'s in 4) mean function composition or matrix multiplication? As I understand it, ##\tau(g)## and ##\rho(g^{-1})## are matrices in ##GL(W)## and ##GL(V)## resp. ?
What is the difference?

fishturtle1
What is the difference?
I think I get it now, I had to look up the definition of GL(V). So, ##\rho(g^{-1})## is an automorphism of ##V## and ##\tau(g)## is an automorphism of ##W## and ##\varphi## is a k linear map from ##V## to ##W##.

Mentor
I think I get it now, I had to look up the definition of GL(V). So, ##\rho(g^{-1})## is an automorphism of ##V## and ##\tau(g)## is an automorphism of ##W## and ##\varphi## is a k linear map from ##V## to ##W##.
Yes. As functions, it is the composition, which in coordinates is matrix multiplication.

fishturtle1
One other question, what is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##? I'm pretty sure ##\text{Hom}_{\mathbb{K}}(V, W)## is the set of all ##\mathbb{K}##-linear maps from ##V## to ##W##. And we can make ##V## into a ##\mathbb{K}G## module by defining ##g \cdot v = \rho(g)v## I think?? So is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, w))## the set of ##\mathbb{K}G## homomorphisms from ##V## to ##W##?

Mentor
One other question, what is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##? I'm pretty sure ##\text{Hom}_{\mathbb{K}}(V, W)## is the set of all ##\mathbb{K}##-linear maps from ##V## to ##W##. And we can make ##V## into a ##\mathbb{K}G## module by defining ##g \cdot v = \rho(g)v## I think?? So is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, w))## the set of ##\mathbb{K}G## homomorphisms from ##V## to ##W##?
I don't see why you need the group field, because linearity of homomorphism spaces is all that is used. ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## means the homomorphisms of representations as defined. We have ##\rho: G \longrightarrow \operatorname{GL}(V)## and ##\tau : G \longrightarrow \operatorname{GL}(W)##. I do not see any functions from ##G## to ##\mathbb{K} .## The condition of the characteristic simply allows us to divide by ##|G|## in the symmetry operator.

fishturtle1
I don't see why you need the group field, because linearity of homomorphism spaces is all that is used. ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## means the homomorphisms of representations as defined. We have ##\rho: G \longrightarrow \operatorname{GL}(V)## and ##\tau : G \longrightarrow \operatorname{GL}(W)##. I do not see any functions from ##G## to ##\mathbb{K} .## The condition of the characteristic simply allows us to divide by ##|G|## in the symmetry operator.
that clears things up, thank you!

\textbf{Claim 1.} ##\text{Sym}(\varphi)## is a linear map from ##V \to W##.

Proof:
First, we have ##\text{char} \mathbb{K} \not\vert \vert G \vert##. So, ##\vert G \vert \neq 0## and ##\frac{1}{\vert G \vert}## is defined. For each ##g \in G##, we have ##(\tau(g) \circ \varphi \circ \rho(g^{-1})(v) \in W##. Since ##W## is a vector space, it is closed under addition and scalar multiplication. So, ##(\text{Sym}\varphi)(v) \in W##. Next, we check that ##\text{Sym}(\varphi)## is ##\mathbb{K}##-linear. Let ##u, v \in V## and ##\lambda \in \mathbb{K}##. We have

\begin{align*}
(\text{Sym}\varphi)(u + v) & = \left(\frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1})\right) (u + v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ \rho(g^{-1}) (u + v)\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ (\rho(g^{-1})(u) + \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ (\varphi \circ \rho(g^{-1})(u) + \varphi \circ \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})(u)) + (\tau(g) \circ + \varphi \circ \rho(g^{-1})(v)) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ \varphi \circ \rho(g^{-1})(u) + \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ + \varphi \circ \rho(g^{-1})(v) \\
&= (\text{Sym}\varphi)(u) + (\text{Sym}\varphi)(v)\\
\end{align*}

Also,

\begin{align*}
(\text{Sym}\varphi)(\lambda v) &= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(\lambda v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \lambda \left(\tau(g) \circ \varphi \circ \rho(g^{-1})(v)\right) \\
&= \lambda \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(v)\right) \\
&= \lambda (\text{Sym}\varphi)(v) \\
\end{align*}

This shows ##\text{Sym}\varphi## is ##\mathbb{K}##-linear.
[]

\textbf{Claim 2.} The map ##\text{Sym}## is a ##\mathbb{K}##-linear mapping.

Proof:
Let ##\varphi, \sigma \in \text{Hom}_{\mathbb{K}}(V, W)## and ##\lambda \in \mathbb{K}##. For any ##g \in G##, we have

\begin{align*}
\text{Sym}(\varphi + \sigma) &= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ (\varphi + \sigma) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ ((\varphi \circ \rho(g^{-1}) + (\sigma \circ \rho(g^{-1}))) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + \frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&= \text{Sym}(\varphi) + \text{Sym}(\sigma)
\end{align*}

Also,

\begin{align*}
\text{Sym}(\lambda\varphi) &= \frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ (\lambda\varphi) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G} k(\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= k\frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= k \text{Sym}(\varphi) \\
\end{align*}

We may conclude ##\text{Sym}## is ##\mathbb{K}##-linear.
[]

\textbf{Claim 3.} ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) = \lbrace \theta : V \longrightarrow W \vert \forall_{g \in G} \tau(g) \circ \theta \circ \rho(g^{-1}) = \theta\rbrace##

Proof:
##(\subseteq)##: Let ##\theta \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. By definition, we have ##\theta \circ \rho(g) = \tau(g) \circ \theta## for all ##g \in G##. In particular, ##\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta \circ \rho(g) \circ \rho(g^{-1}) = \theta##.
\\

##(\supseteq)##: Suppose for all ##g \in G##, ##\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta##. Multiplying both sides by ##\rho(g)##, we have ##\tau(g) \circ \theta = \theta \circ \rho(g)##. This shows ##\supseteq##.
[]

\textbf{Claim 4.} ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## is a subspace of ##\text{Hom}_{\mathbb{K}}(V, W)##.

Proof:
Consider the map that sends every element in ##V## to ##0##. This map is contained in ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. So, ##\emptyset \neq \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) \subset \text{Hom}_{\mathbb{K}}(V,W)##. Let ##\varphi, \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)), \lambda \in \mathbb{K}## and ##g \in G##.
\\

We have $$(\varphi + \sigma)\circ \rho(g) = (\varphi \circ \rho(g)) + (\sigma\circ \rho(g)) = (\tau(g) \circ \varphi) + (\tau(g) \circ \sigma) = \tau(g)(\varphi + \sigma)$$
So, ##\varphi + \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. Also,
$$((\lambda \varphi) \circ \rho(g)) = \lambda(\varphi \circ \rho(g)) = \lambda (\tau(g) \circ \varphi) = (\lambda\tau(g)) \circ \varphi = \tau(g) \circ (\lambda \varphi)$$

So, ##\lambda \varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. We can conclude ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## is a subspace of ##\text{Hom}_{\mathbb{K}}(V, W)##.

[]

\textbf{Claim 5.} ##\text{Sym}## is a projection onto ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##.

Proof:
Let ##\varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. Then ##\text{Sym}(\varphi) = \varphi##. In particular, ##\text{Sym}(\text{Sym}(\varphi)) = \text{Sym}(\varphi)##. It follows that ##\text{Sym}^2 = \text{Sym}## on ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. This proves Claim 5.
[]

Mentor
problem 4
\textbf{Claim 1.} ##\text{Sym}(\varphi)## is a linear map from ##V \to W##.
(1) ##\operatorname{Im(Sym)} \subseteq \operatorname{Hom}(V,W).## You haven't pointed out that well-definition is one of the 5 claims, as has been already mentioned by @Infrared. But since you mentioned it implicitly in your proof, I take the following for well-definition, too. (2)
Proof:
First, we have ##\text{char} \mathbb{K} \not\vert \vert G \vert##. So, ##\vert G \vert \neq 0## and ##\frac{1}{\vert G \vert}## is defined. For each ##g \in G##, we have ##(\tau(g) \circ \varphi \circ \rho(g^{-1})(v) \in W##. Since ##W## is a vector space, it is closed under addition and scalar multiplication. So, ##(\text{Sym}\varphi)(v) \in W##. Next, we check that ##\text{Sym}(\varphi)## is ##\mathbb{K}##-linear. Let ##u, v \in V## and ##\lambda \in \mathbb{K}##. We have

\begin{align*}
(\text{Sym}\varphi)(u + v) & = \left(\frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1})\right) (u + v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ \rho(g^{-1}) (u + v)\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ (\rho(g^{-1})(u) + \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ (\varphi \circ \rho(g^{-1})(u) + \varphi \circ \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})(u)) + (\tau(g) \circ + \varphi \circ \rho(g^{-1})(v)) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ \varphi \circ \rho(g^{-1})(u) + \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ + \varphi \circ \rho(g^{-1})(v) \\
&= (\text{Sym}\varphi)(u) + (\text{Sym}\varphi)(v)\\
\end{align*}

Also,

\begin{align*}
(\text{Sym}\varphi)(\lambda v) &= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(\lambda v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \lambda \left(\tau(g) \circ \varphi \circ \rho(g^{-1})(v)\right) \\
&= \lambda \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(v)\right) \\
&= \lambda (\text{Sym}\varphi)(v) \\
\end{align*}

This shows ##\text{Sym}\varphi## is ##\mathbb{K}##-linear.
[]

\textbf{Claim 2.} The map ##\text{Sym}## is a ##\mathbb{K}##-linear mapping.
Linearity. (3) This is basically clear from the definition.
Proof:
Let ##\varphi, \sigma \in \text{Hom}_{\mathbb{K}}(V, W)## and ##\lambda \in \mathbb{K}##. For any ##g \in G##, we have

\begin{align*}
\text{Sym}(\varphi + \sigma) &= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ (\varphi + \sigma) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ ((\varphi \circ \rho(g^{-1}) + (\sigma \circ \rho(g^{-1}))) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + \frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&= \text{Sym}(\varphi) + \text{Sym}(\sigma)
\end{align*}

Also,

\begin{align*}
\text{Sym}(\lambda\varphi) &= \frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ (\lambda\varphi) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G} k(\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= k\frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= k \text{Sym}(\varphi) \\
\end{align*}

We may conclude ##\text{Sym}## is ##\mathbb{K}##-linear.
[]

\textbf{Claim 3.} ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) = \lbrace \theta : V \longrightarrow W \vert \forall_{g \in G} \tau(g) \circ \theta \circ \rho(g^{-1}) = \theta\rbrace##
This is no claim. It is the definition of a homomorphism of representations. You have to prove that it holds for the symmetry operator as we defined it, i.e. that all ##\operatorname{Sym}(\varphi )## "commute" with the representations. It is actually one of two points where an argument is necessary. (The projection is the other one.)
Proof:
##(\subseteq)##: Let ##\theta \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. By definition, we have ##\theta \circ \rho(g) = \tau(g) \circ \theta## for all ##g \in G##. In particular, ##\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta \circ \rho(g) \circ \rho(g^{-1}) = \theta##.
\\

##(\supseteq)##: Suppose for all ##g \in G##, ##\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta##. Multiplying both sides by ##\rho(g)##, we have ##\tau(g) \circ \theta = \theta \circ \rho(g)##. This shows ##\supseteq##.
[]

\textbf{Claim 4.} ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## is a subspace of ##\text{Hom}_{\mathbb{K}}(V, W)##.
I actually did not count this as a claim, since the linear spaces are already included by definition. We have some homomorphisms with an additional condition
$$\{\vartheta :V\longrightarrow W\,|\,\forall_{g\in G}\, : \,\tau(g)\circ\vartheta\circ \rho(g^{-1})=\vartheta \}$$
and all homomorphisms ##\operatorname{Hom}(V,W)## on the other hand. That it is a subspace follows from the linearity in the condition, which is already contained in your claim (2) if you drop the sums.
Proof:
Consider the map that sends every element in ##V## to ##0##. This map is contained in ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. So, ##\emptyset \neq \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) \subset \text{Hom}_{\mathbb{K}}(V,W)##. Let ##\varphi, \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)), \lambda \in \mathbb{K}## and ##g \in G##.
\\

We have $$(\varphi + \sigma)\circ \rho(g) = (\varphi \circ \rho(g)) + (\sigma\circ \rho(g)) = (\tau(g) \circ \varphi) + (\tau(g) \circ \sigma) = \tau(g)(\varphi + \sigma)$$
So, ##\varphi + \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. Also,
$$((\lambda \varphi) \circ \rho(g)) = \lambda(\varphi \circ \rho(g)) = \lambda (\tau(g) \circ \varphi) = (\lambda\tau(g)) \circ \varphi = \tau(g) \circ (\lambda \varphi)$$

So, ##\lambda \varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. We can conclude ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## is a subspace of ##\text{Hom}_{\mathbb{K}}(V, W)##.

[]

\textbf{Claim 5.} ##\text{Sym}## is a projection onto ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##.

Proof:
Let ##\varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. Then ##\text{Sym}(\varphi) = \varphi##. In particular, ##\text{Sym}(\text{Sym}(\varphi)) = \text{Sym}(\varphi)##. It follows that ##\text{Sym}^2 = \text{Sym}## on ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. This proves Claim 5.
[]
This is wrong. ##\operatorname{Sym}(\varphi ) \stackrel{i.g.}{\neq } \varphi .## You have to calculate that ##\operatorname{Sym}^2(\varphi )=\operatorname{Sym}(\varphi )##.

You missed both crucial points which actually require some "proof":
• ##\tau(h)\circ\operatorname{Sym}(\varphi)=\operatorname{Sym}(\varphi)\circ \rho(h) ##
• ##\operatorname{Sym}^2(\varphi )=\operatorname{Sym}(\varphi )##

fishturtle1
Mentor
Correction to the last point: I overlooked that you have chosen ##\varphi \in \operatorname{Hom}_\mathbb{K}((\rho,V),(\tau,W))##. So ##\operatorname{Sym}(\varphi )=\varphi ## indeed, but why?

Correction to the last point: I overlooked that you have chosen ##\varphi \in \operatorname{Hom}_\mathbb{K}((\rho,V),(\tau,W))##. So ##\operatorname{Sym}(\varphi )=\varphi ## indeed, but why?
I think the calculation is

\begin{align*}
\text{Sym}(\varphi) &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \\
&= \frac{1}{\vert G \vert} \cdot \vert G \vert \varphi \\
&= \varphi \\
\end{align*}

Edit: Also, thank you for the feedback and corrections.

fresh_42
Mentor
I think the calculation is

\begin{align*}
\text{Sym}(\varphi) &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \\
&= \frac{1}{\vert G \vert} \cdot \vert G \vert \varphi \\
&= \varphi \\
\end{align*}

Edit: Also, thank you for the feedback and corrections.
Now for the last one: why does ##\operatorname{Sym}(\varphi )## commute with representation matrices?

fishturtle1
For ##\varphi \in \text{Hom}_{\mathbb{K}} ((\rho, V), (\tau, W))##, we have ##\tau(h) \text{Sym}(\varphi) = \text{Sym}(\varphi) \rho(h)##

Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h) \circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ (\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \varphi \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) \circ \rho(h) \\
&= \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))\right) \circ \rho(h) \\
&= \text{Sym}\varphi \circ \rho(h) \\
\end{align*}

Mentor
For ##\varphi \in \text{Hom}_{\mathbb{K}} ((\rho, V), (\tau, W))##, we have ##\tau(h) \text{Sym}(\varphi) = \text{Sym}(\varphi) \rho(h)##

Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h) \circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ (\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \varphi \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) \circ \rho(h) \\
&= \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))\right) \circ \rho(h) \\
&= \text{Sym}\varphi \circ \rho(h) \\
\end{align*}
Yes, but ##\operatorname{Sym}\, : \,\operatorname{Hom}(V,W)\longrightarrow \operatorname{Hom}(V,W)## and we want to show that actually ##\operatorname{Sym}\, : \,\operatorname{Hom}(V,W)\longrightarrow \operatorname{Hom}((\rho,V),(\tau,W))##. That is, we have an arbitrary homomorphism ##\varphi \, : \,V\longrightarrow W##, and only its image satisfies the additional condition, which we want to show.

##\operatorname{Sym}## is a projection, i.e. maps something from bigger to smaller.

Hint: We haven't used that ##\rho## and ##\tau## are representations, yet.

fishturtle1
Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h)\circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(h) \circ \tau(h^{-1}g) \varphi \circ \rho(g^{-1}h) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \text{Sym}(\varphi) \circ \rho(h) \\
\end{align*}

In the above calculations, we used that ##\rho## and ##\tau## are homomorphisms.

fresh_42
Mentor
Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h)\circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(h) \circ \tau(h^{-1}g) \varphi \circ \rho(g^{-1}h) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \text{Sym}(\varphi) \circ \rho(h) \\
\end{align*}

In the above calculations, we used that ##\rho## and ##\tau## are homomorphisms.
... and the fact that left multiplication in a group is a bijection ##L_{h^{-1}}\, : \,g \longmapsto h^{-1}g##.

Just saying, because the difficulty of the problem was mainly to identify all those seemingly clear facts. It is sometimes more difficult to see what has to be shown than it is to show it.

fishturtle1