Is any representation equivalent to a unitary one? I work in a compact or finite group.
First question (just to check) :
Can I say that any representation is equivalent to a unitary one ?
And as the unitary representations are always completly reducible, thus any representation of a finite or compact group is completely reducible.
I have seen the equivalence property somewhere but I don't find a proof for it (and I would check with you if it is true indeed).
Second question : (if the first is true) :
Is there an easy proof for "a representation is equivalent to a unitary one" for compact groups ? I know some linear algebra (but I am not very very familiar with Hilbert space from a rigorous math point of view).
 A: It seems like you are not totally clear about the two representations being equivalent and the meaning of a unitary representation.
Equivalent representations

Let $V$ and $W$ be representations of a group $G$. A morphism $f:V\to W$ is a linear and equivariant map, i.e., for all $g \in G$ and $v
\in V$ $f(gv)=gf(v)$ (basically morphisms of vectorspaces or in
  general of $G-$modules). If such a morphism has an inverse we call it
  an isomorphism and the representations $V$ and $W$ equivalent.

Unitary representation

Let $G$ be a compact group and $V$ a representation of $G$. An inner
  product $V \times V \to \mathbb C$ is called $G-$invariant if $\langle
gu,gv \rangle = \langle u,v \rangle$ holds for all $g \in G$ and for
  all $v,u \in V$. A representation together with a $G-$invariant inner
  product is called a unitary representation.

The following fact is not hard to prove: Every representation $V$ of a compact group possesses a $G-$invariant inner product.
Let $b:V\times V \to \mathbb C$ be any inner product and define $$c(u,v) = \int_G~b(gu,gv) \operatorname{d}g$$ where the integral is normalized and left-invariant (this is a so called Haar-integral which exists on every compact group and which is unique). Let's check if $c$ defines a $G-$invariant inner product: $c$ is linear in $u$ and conjugate linear in $v$ (since $b$ is a (hermitian) inner product). $c$ is $G-$invariant since the integral is left-invariant and $c$ is positive definite since the integral of a positive continuous function is positive. 
Therefore $c$ is a $G-$invariant inner product and $V$ together with $c$ a unitary representation.
In the case of finite groups you can use sums instead of the integral and everything should work in a very similar matter.
A: I think yes, assuming the characteristic is zero. Indeed you can deform a scalar product $(-,-)$ into a $G$-invariant scalar product $\langle -,- \rangle$ if $G$ is compact or finite, defining $\langle x, y \rangle = \sum_{g \in G} (gx,gy)$. Your new scalar product is preserved by $G$, thus chaning of basis you get an unitary representation.
If $G$ is compact you should integrate instead of summing, but really it is completely elementary and exactly the same idea. As you said this is indeed the key step for proving Maschke's theorem (again, assuming the characteristic is zero), because if $U$ is $G$-invariant, then $U^{\perp}$ is a $G$-invariant complementary subspace. 
