On irreducible complex representations I have to prove this:
If $V$ is an irreducible complex representation of the group $G$ then $End(V)$ is generated by the elements of $G$.
I can't figure out how to solve this problem.
I don't know if this problem can be' solved without any specific knowledge of representation theory, I only tried with standard linear algebra arguments.
 A: As pointed out in the comments, it follows from the Jacobson density theorem, which has arguably elementary proofs.
But here's an argument using only "basic" linear algebra.
Let $A\subset End(V)$ be an irreducible subalgebra. That is, a subalgebra such that $V$ has no nontrivial subspaces that are stable under the action of $A$.
I claim that $A=End(V)$. If we manage to prove this, then letting $A$ be the subalgebra generated by $G$, we get the result you're after.
Now, to prove this claim. Let me first note that if $A$ contains one rank $1$ operator, then it contains all of them.
Indeed, a rank one operator is of the form $v\mapsto l(v) w$ for some $l\in V^*, w\in V$, which I'll denote by $l\otimes w$. Then note that $A$ acts on $V^*$ the right via $(l,a)\mapsto l\circ a$, and this action is irreducible as well : if $F\subset V^*$ is stable under $A$, then $F^\bot = \{v\in V\mid \forall l\in F, l(v)=0\}$ is stable under $A$ in $V$, and it's nontrivial precisely if $F$ is.
In particular if $A$ contains $l\otimes v$, it contains $l\otimes av$ for all $a\in A$ (this is just $a\circ (l\otimes v)$) and it also contains $la \otimes v$ (this is $(l\otimes v)\circ a$).
Since $Av = V$ (otherwise it would be a nontrivial stable subspace) and $lA = V^*$ (for the same reason), we get that if $A$ contains $l\otimes v$, it contains $l'\otimes v'$ for any $l'\in V^*, v'\in V$, i.e. it contains all rank $1$ operators. Hence, since those span $End(V)$, if $A$ contains a single rank $1$ operator, it contains all of them.
We therefore simply need to prove that $A$ contains at least a rank $1$ operator. For this we proceed as follows : we let $f\in A$ have rank $>1$ and prove that we can find a nonzero $f'\in A$ with $rank(f')<rank(f)$. This will prove the claim.
This is where we use that we're over $\mathbb C$ (or more generally, an algebraically closed field).
Let $x,y$ be such that $(f(x),f(y))$ are linearly independent (this exists by assumption on $rank(f)$). Then there is $g\in A$ such that $gf(x) = y$. Consider then the endomorphism $fg$ of $\mathrm{im}(f)$ : it has an eigenvalue $\lambda$.
I claim that $fgf-\lambda f$ is nonzero but has rank $<rank(f)$.
First, the rank : our morphism is $(fg-\lambda id_V)\circ f$, and $fg-\lambda id_V$ has nonzero kernel on $\mathrm{im}(f)$, so the rank is $<f$.
Second, it's nonzero: indeed evaluate it on $x$, you get $f(y) - \lambda f(x)\neq 0$ by hypothesis.
Since it's in $A$, we got what we wanted; so we get an element in $A$ of rank $1$, so we get all of them, so we get all of $End(V)$
