How to use a proof by contradiction in group theory! A group $G$  is said to be abelian-by-finite if it has a normal subgroup $H$ of finite index in $G$.
I want to prove the following statement: "If a group $G$ has property $\mathcal P$ then it is  abelian-by-finite", and I want to use the proof by contradiction.
Can I formulate it in this way: Suppose $G$ has property $\mathcal P$ and it is not abelian-by-finite, therefore it exists a normal subgroup $H$ which is not of finite index in $G$? Can I use this to prove this statement by contradiction?
 A: We want to prove $A \implies B$ where $A$ is "abelian-by-finite" and $B$ is property $P$.  To prove this by contradiction you could assume $A$ and $\neg B$ and derive a contradiction.  Or you could prove the contrapositive $\neg B \implies \neg A$, which is equivalent.  Either way, assuming $B$, that $G$ has property $P$, will not help you as it is the thing you are trying to prove. 
A: You want to prove the implication
$$
 \text{$G$ has property $\mathcal{P}$}
 \implies
 \text{$G$ is abelian-by-finite},
$$
If you want to do a proof by contradiction you have to assume that $G$ has property $\mathcal{P}$ (i.e. that the left hand side holds) but that $G$ is not abelian-by-finite (i.e. that the right hand side does not hold).
Imagine  it like this: If you want to prove the implication
$$
 \text{it rains} \implies \text{the street gets wet}
$$
by contradiction then you assume that it rains, but that the street stays dry, and show that this leads to a contradiction.

Regarding your formulation: The converse of $G$ being abelian-by-finite is that $G$ has no normal subgroup of finite index, which is not equivalent to $G$ having a normal subgroup of infinite index (every infinite group has a normal subgroup of infinite index, namely the trivial subgroup, but an infinite group can still be abelian-by-finite). Instead it is equivalent to every normal subgroup of $G$ having infinite index.
(In your definition of abelian-at-finite you probably want to exclude $G$ itself as a normal subgroup, unless you want every group to be abelian-at-finite.)
A: It was already said that having an infinite normal sub Group dos not mean that there is not a finite one. I wanted to elaborate on that with an example:
Consider A = the Boolean Group(also filed) under addition(tats 1,0 where 1+1=0).
And B = the Group of 2*2 matrix's under R.
Now let us look at A*B. This new Group(of Pairs) have an infinite normal subgroup:
{(1,a*I) | a in R} for example.
but it also have a finite normal Group:
{(a,0) | a in A}
(I promise that today I will learn how to use mathematical notations..)
A: 
Can I formulate it in this way: Suppose $G$ has property $\mathcal P$ and it is not abelian-by-finite, therefore it exists a normal subgroup $H$ which is not of finite index in $G$?

No.
The statement you're trying to negate is "$G$ has a normal subgroup $H$ of finite index in $G$."
That means there exists such a subgroup $H$.
The negation of $$\text{“There exists a group } H \text{ such that } \varphi(H)\text{''}$$ is $$\text{“For every group } H \text{ not } \varphi(H).\text{''}$$
The negation of
$$
\text{“There exists a subgroup } H \text{ of } G \text{ such that } H \text{ has finite index in } G \text{''}
$$
is
$$
\text{“Every subgroup } H \text{ of } G \text{ has infinite index in } G. \text{''}
$$
Every subgroup, not just one subgroup.
