I'm stuck with the following statement.
Let $G$ be a group, $H$ a normal subgroup of $G$ and $G/H$ the quotient group. If $H$ and $G/H$ are both torsion-free, then $G$ is also torsion-free.
I think it's true by intuition and I believe this follows from the fact that you can unify the equivalence classes which are torsion-free because $G/H$ is and in that way obtain $G$.
Nevertheless I'm not able to give an exact proof of it and would be really thankful for a good advice or a short proof!
Let $x$ be a nontrivial element in $G$.
Suppose that $x$ is of finite order, say $n$.
Then $x^n=1$.
Hence $(xH)^n=x^nH=H$.
This means that $xH$ is element of finite order in $G/H$.
Since $G/H$ is torsion-free, $xH=H$, so $x\in H$.
But this is a contradiction because $H$ is also torsion-free.
Thus we conclude that $G$ is torsion-free.