Direct sum and tensor product of two representations of a group Our lecturer gave us a hard exercice to go further in group theory (we stopped at group actions) :
Let G be a group, V and W complex vector spaces and $\rho_1 : G \mapsto GL(V) $ be a group homomorphism where GL(V) is the general linear group of V 
(i.e. the invertible linear maps V $\mapsto$ V, group under compositions of maps) and let
$\rho_2 : G \mapsto GL(W) $.
$(\rho_1,V)$ and $(\rho_2,V)$ are called representation of G. I have to show that :
-For representations V, W of G, the direct sum $V\oplus W$ is a representation of G.
i.e. I have to find a homomorphism $\omega_1 :  G \mapsto GL(V\oplus W)$.
-Same again but for the tensor product $V\otimes W$ via $g(v\otimes w):=gv\otimes gw$
i.e. I have to find a homomorphism $\omega_2 : G \mapsto GL(V\otimes W)$.
The problem is : I struggle with the definitions of direct sum and tensor product... (never worked on this before).
Here is what I've done for the first question.
 I have to show that : let $v\in V$ and $w\in V$ and $(\rho, V\oplus W)$ be a representation of G. Then
$\forall g\in G, \rho(g)(v+w)=\rho(g)(v)+\rho(g)(w):=\rho_1(g)(v)+\rho_2(g)(w) \in V\oplus W$. But I don't know how to prove it.
I tried to find lectures and understand as much as I could but there is usually no explanations about those two questions, it is considered trivial apparently !
Thanks for your help
 A: *

*Notation
$(G,\cdot_G)$ is a group with composition (or product) $\cdot_G$. The group of automorphisms of a vector space, let us say $V$, is denoted by $(\operatorname{Aut}(V),\circ)$, where $\circ$ is the composition of automorphisms.
The representations are defined in the OP; we use the following notation
$$\rho_1: (G,\cdot_G)\rightarrow (\operatorname{Aut}(V),\circ),$$
$$\rho_2:(G,\cdot_G)\rightarrow (\operatorname{Aut}(W),\circ).$$
 We just recall that given any representation $\rho: (G,\cdot_G)\rightarrow (\operatorname{Aut}(T),\circ)$ we have
$$\rho(g_1\cdot_G g_2)=\rho(g_1)\circ\rho(g_2)$$
for all $g_1,g_2\in G$.


*

*On direct sum.


$$ \rho_\oplus:(G,\cdot_G)\rightarrow (\operatorname{Aut}(V\oplus W),\circ)$$
is given by $\rho_\oplus:=\rho_1\oplus\rho_2$, i.e. 
$$\rho_\oplus(g)(v\oplus w)=\rho_1(g)(v)\oplus\rho_2(g)(w)\in V\oplus W$$ for all $v \in V$, $w\in W$ and $g\in G$.
By definition, it follows that $\rho_\oplus(g_1\cdot_G g_2)=\rho_\oplus(g_1)\circ\rho_\oplus(g_2)$ and $\rho_\oplus(g^{-1})=\rho^{-1}_\oplus(g)$. This makes $\rho_\oplus$ a group homomorphism. Let us prove the first one as example. The first equation is proven by
$$(\rho_\oplus(g_1\cdot_G g_2))(v\oplus w)=(\text{def. of}~\rho_\oplus)=
\rho_1(g_1\cdot_G g_2)(v)\oplus\rho_2(g_1\cdot_G g_2)(w)=(\text{def. of representations:})=
\rho_1(g_1)(\rho_1(g_2)(v))\oplus\rho_2(g_1)(\rho_2(g_2)(w))=
(\rho_\oplus(g_1)\circ\rho_\oplus(g_2))(v\oplus w);$$
the last equality follows from the definition of composition $\circ$ in $\operatorname{Aut}(V\oplus W)$. In fact:
$$(\rho_\oplus(g_1)\circ\rho_\oplus(g_2))(v\oplus w):=
\rho_\oplus(g_1)(\rho_\oplus(g_2)(v\oplus w))=(\text{def. of}~\rho_\oplus)=
\rho_\oplus(g_1)(\underbrace{\rho_1(g_2)(v)}_{\in V}\oplus \underbrace{\rho_2(g_2)(w)}_{\in W})=(\text{again def. of}~\rho_\oplus)=\\
\underbrace{\rho_1(g_1)(\rho_1(g_2)(v))}_{\in V}\oplus \underbrace{\rho_2(g_1)(\rho_2(g_2)(w))}_{\in W},$$
as wished.


*

*On tensor product


$$ \rho_\otimes:(G,\cdot_G)\rightarrow (\operatorname{Aut}(V\otimes W),\circ)$$
is given by $\rho_\otimes:=\rho_1\otimes\rho_2$, i.e. 
$$\rho_\otimes(g)(v\otimes w)=\rho_1(g)(v)\otimes\rho_2(g)(w)\in V\otimes W$$ for all $v \in V$, $w\in W$ and $g\in G$.
By definition, it follows that $\rho_\otimes(g_1\cdot_G g_2)=\rho_\otimes(g_1)\circ\rho_\otimes(g_2)$ and $\rho_\otimes(g^{-1})=\rho^{-1}_\otimes(g)$. This makes $\rho_\otimes$ a group homomorphism. The relations are proven in a similar way to the one used in the direct sum case.
