This does not work : why is $f$ equivariant ?
Given representations $\rho_1,\rho_2$ of $G$ on $V$, $id_V: (V,\rho_1)\to (V,\rho_2)$ is equivariant if and only if $\rho_1=\rho_2$ : this has nothing to do with Schur's lemma, or irreducibility, this is actually very easy to see.
Now you claim that $id$ is equivariant but to prove that you'd have to first prove that $\Gamma = \rho_1\otimes \rho_2$, which is what you want to prove anyway.
Note that your proof can't work anyway because you chose arbitrary $\rho_1,\rho_2$, so it's clear that it can't work.
Martin Brandenburg's answer in your related question proves this fact (he mentions it in his answer) in the case of an algebraically closed field of characteristic prime to $|G_1|,|G_2|$; if you can prove that $\rho_1\otimes \rho_2\cong \rho'_1\otimes \rho'_2 \implies \rho_1\cong \rho'_1 \land \rho_2\cong \rho'_2$.
However, Mariano Suarez-Alvarez already gives a proof (still in your related question) of your statement : he actually produces two irreducible representations $V_1,V_2$ of $G_1,G_2$ respectively whose tensor product is isomorphic to the $\Gamma$ you start with - you should definitely check out his answer, it's well-written !
The only point in it that is not entirely clear at a first glance is why $\hom_G(U,V_{\mid G}) $ has dimension $\leq \frac{\dim V}{\dim U}$, but that follows at once by using Schur's lemma and decomposing $V_{\mid G}$ into irreducible representations
(although to make sure that that holds you probably have to use some hypothesis, again something like the fact that the field is algebraically closed of characteristic prime to $|G|$)
Firstly, there is a general version of Schur's lemma that only relies on irreducibility, and not on any finiteness conditions or anything about the base field.
(Schur's lemma, general) Let $G$ be a group, and $f: V \to W$ a $G$-equivariant map between representations. Then:
- If $V$ is irreducible, then $f$ is either zero or injective,
- If $W$ is irreducible, then $f$ is either zero or surjective,
- If both $V$ and $W$ are irreducible, then $f$ is either zero or an isomorphism.
The proof is almost by definition of an irreducible representation. For example, by $G$-equivariance of $f$, the subspace $\ker f \subseteq V$ is a $G$-stable subspace, and then in (1) by irreducibility of $V$, we must either have $\ker f = V$ (hence $f$ is zero) or $\ker f = 0$ (hence $f$ is injective). The proofs of (2) and (3) are similar.
The next version of Schur's lemma holds for finite-dimensional irreducible representations over an algebraically closed field. $G$ can be infinite, but also note that if $G$ is finite, then any irreducible representation is automatically finite-dimensional.
(Schur's lemma, algebraically closed field) Let $G$ be a group, $V$ an irreducible representation of $G$ over the algebraically closed field $k$, and $f: V \to V$ a $G$-equivariant endomorphism. Then $f$ is a scalar map, i.e. $f = \lambda \operatorname{id}_V$ for some $\lambda \in k$.
The proof goes: since $k$ is algebraically closed and $V$ is finite-dimensional, $f$ has some eigenvalue $\lambda \in k$. The linear map $f - \lambda \operatorname{id}_V$ is then a $G$-equivariant map between irreducible representations, but it is not invertible, so it is zero by the general version of Schur's lemma. Hence $f = \lambda \operatorname{id}_V$.
The last result about abelian groups follows from the algebraically closed version of Schur's lemma. Again it needs no finiteness conditions on $G$.
Let $G$ be an abelian group, and $V$ a finite-dimensional irreducible representation of $G$ over an algebraically closed field. Then $V$ is one-dimensional.
The proof goes like this. For any $g \in G$ we can define the left-multiplication map $L_g: V \to V$ by $L_g(v) = gv$. Since $G$ is abelian, $L_g$ is a $G$-equivariant map: for any $h \in G$ we have $L_g(hv) = ghv = hgv = h L_g(v)$. Applying the algebraically closed version of Schur's lemma, we see that $L_g$ is a scalar, and therefore every element of $G$ just scales vectors. So for any nonzero $v \in V$, the subspace spanned by $v$ is a subrepresentation, and by irreducibility must be the whole space.
Best Answer
There are many types of homomorphisms/isomorphisms in mathematics. Here the meaning is not an isomorphism of groups, but an isomorphism of representations. Given two representations $(V_1,\pi_1), (V_2,\pi_2)$ an isomorphism is an invertible linear map $A:V_1\to V_2$ such that $A(g.v)=g.(Av)$ for all $g\in G, v\in V_1$.