First some definitions:
A linear representation of a group $G$ is a group homomorphism $\rho:G\to GL(V)$, where $GL(V)$ is the group consisting of all isomorphisms on the linear space $V$, equipped with the operation of composition.
Two representations $\rho_1:G\to GL(V_1)$ and $\rho_2:G\to GL(V_2)$ are said to be similar if $V_1$ and $V_2$ are isomorphic and there exists an isomorphism $\tau: V_1\to V_2$ such that $\rho_1(g)\circ\tau=\tau\circ\rho_2(g)$ for every $g\in G$. This is denoted by $\rho_1\simeq\rho_2$.
A representation is said to be finite-dimensional if $V$ is finite-dimensional.
If $\rho_1:G\to GL(V_1)$ and $\rho_2:G\to GL(V_2)$ are linear representations, then we define the direct sum of $\rho_1$ and $\rho_2$ to be the function $\rho_1\oplus\rho_2:G\to V_1\oplus V_2$ given by $(\rho_1\oplus\rho_2(g))(v_1+v_2) = (\rho_1(g))(v_1)+(\rho_2(g))(v_2)$.
Now I'm supposed to show this:
Let $\rho_1 \simeq \rho_2$ and $\rho_3 \simeq \rho_4$ be finite-dimensional linear representations of a finite group $G$. Then $\rho_1\oplus\rho_3\simeq\rho_2\oplus\rho_4$.
I tried to to translate $\rho_1(g)\circ\tau=\tau\circ\rho_2(g)$ into the matrix equality but I'm stuck in getting anything from this, can anyone help? Or maybe I need more theory to answer this?
Best Answer
Surely this $\tau:V_1\to V_2$ in your definition of "similar" should be a (linear) isomorphism as well. Otherwise any two representations with the same dimension would be similar (note that the trivial linear map $\tau(x)=0$ always satisfies the similarity condition). And thus your claim would be trivially true.
Anyway, by the assumption we have a representation isomorphisms $\tau:V_1\to V_2$ and $\theta:V_3\to V_4$. Those linear isomorphisms induce a linear map:
$$\gamma:V_1\oplus V_3\to V_2\oplus V_4$$ $$\gamma(x,y)=\big(\tau(x), \theta(y)\big)$$
It is easy to see that it is a linear isomorphism (the inverse is of the same form).
It also satisfies the conditions for similarity:
$$(\rho_1\oplus\rho_3)(\gamma(x,y))=(\rho_1\oplus\rho_3)(\tau(x),\theta(y))=$$ $$=\rho_1(\tau(x))\oplus \rho_3(\theta(y))=\tau(\rho_2(x))\oplus\theta(\rho_4(y))=$$ $$=(\tau\oplus\theta)\big((\rho_2\oplus\rho_4)(x,y)\big)=\gamma\big((\rho_2\oplus\rho_4)(x,y)\big)$$
and thus
$$(\rho_1\oplus\rho_3)\circ\gamma=\gamma\circ(\rho_2\oplus\rho_4)$$
which completes the proof.
Note that $V_i$ being of finite dimension is irrelevant.