if $N$ and $G/N$ are torsion-free groups, what can we say of $G$? My question is kind of selly, but I really need to understand the consept very well, so my problem is:
Let $G$ be a torsion-free group (i.e., each nontrivial $x \in G$ has an infinite order).
I know that if $H \subset G$ then it is also torsion-free, and if $N  \trianglelefteq G$ then $G/N$ is not torsion-free.
I would know if $N$ and $G/N$ are both torsion- free groups, then is $G$ also a torsion-free group or not?
And if $G$ is a torsion-free group, is it of infinite order?

Let us clear up a few things.
A group $G$ is torsion-free if and only if, for all $x \in G$, either $x$ is the identity or $x$ has an infinite order (beware the identity!). In particular, the trivial group is torsion-free.
Let $G$ be a torsion-free group. Either $G$ is trivial, or $G$ has an element $x$ which is not the identity. In the later case, the sequence $(x^n)_{n \in \mathbb{Z}}$ is a sequence of pairwise distinct elements of $G$, hence $G$ is of infinite order. So you have the dichotomy: either $G$ is trivial, or it has infinite order.
Let $G$ be torsion-free, and $N < G$. Then $G/N$ may or may not be torsion-free. For instance, 


*

*take $G := \mathbb{Z}$ and $N := \{0\}$. Then $G/N \simeq \mathbb{Z}$ is torsion-free.

*take $G := \mathbb{Z}$ and $N := 2 \mathbb{Z}$. Then $G/N \simeq \mathbb{Z}/2\mathbb{Z}$ is not torsion-free.

Finally, let me answer your question. Le $G$ be a group, $N < G$ a normal subgroup of $G$. Assume that $N$ and $G/N$ are torsion-free. We want to prove that $G$ is also torsion-free.
Let $\pi : G \to G/N$, $\pi (x) = xN$ be the canonical morphism.
Let $x \in G$. If $x$ has finite order, then there exists $n > 0$ such that $x^n = e$, so $N = \pi(x^n) = \pi(x)^n$. Hence, $\pi(x)$ also has finite order in $G/N$. Since $G/N$ is assumed to be torsion-free, its only element with finite order is the identity, so $\pi(x) = N$, that is, $x \in N$. Since $N$ is also assumed to be torsion-free, its only element with finite order is $e$, so $x = e$.
Hence, the only element of $G$ with finite order is its identity, so $G$ is also torsion-free.
