$0 \to G' \xrightarrow{\alpha} G \xrightarrow{\beta} G'' \to 0$ exact sequence, $G$ is torsion group if and only if $G'$ and $G''$ are torsion groups. Let's work in abelian groups. We say a group $G$ is a torsion group if for every $g \in G$ there is a $n \geq 0$ such as $ng=0$. I want to prove that for an exact sequence
$$0 \to G' \xrightarrow{\alpha} G \xrightarrow{\beta} G'' \to 0,$$
$G$ is a torsion group if and only if $G'$ and $G''$ are torsion groups.
I cant prove this problem so I formulate an easier but similar problem.
Let's take the particular exact sequence
$$0 \to G' \xrightarrow{i} G \xrightarrow{\pi} \frac{G}{G'} \to 0,$$
where $i$ is the inclusion of the subgroup $G'$ of $G$ and $\pi$ is a projection. If $G$ is a torsion  group then it is inmediate $G'$ is also a torsion group. Also if $g+G' \in \frac{G}{G'}$,and as $g \in G$ where $G$ is a torsion group then there is a $n \geq 0$ such $ng=0$. Then $n(g+ G')=ng +G'=0+G$, proving $\frac{G}{G'}$ is a torsion group.
If $G'$ and $\frac{G}{G'}$ are torsion groups and lets take $g \in G$, then as $\pi(g)=g+G' \in \frac{G}{G'}$ so there is a $n \geq 0$ such $n(g+G')=ng+G'=0$, so $ng \in G'$ and $G'$ is torsion group so there is a $n' \geq 0$ such $n'(ng)==(n'n)g=0$. Proving $G$ is a torsion group.
Still cant generalize this problem to te original one. Any help solving the original proof with this or other aproach will be apreciated.
 A: Let $1\to G’\stackrel{\alpha}{\to}G\stackrel{\beta}{\to}G’’\to 1$ be an exact sequence of (abelian) groups. We want to show that $G$ is torsion if and only if both $G’$ and $G’’$ are torsion.
If $G$ is torsion, let $x\in G’$. Then $\alpha(x)$ is torsion, so there exists $n\gt 0$ such that $\alpha(x)^n = e$; then $\alpha(x^n)=e$, so $x^n\in\mathrm{ker}(\alpha)$. Therefore, $x^n=e$. This shows every element of $G’$ is torsion. If $y\in G’’$, then we know there exists $g\in G$ such that $\beta(g)=y$. Since $G$ is torsion, there exists $m\gt 0$ such that $g^m=e$. Then $y^m = \beta(g)^m = \beta(g^m) = \beta(e)= e$. Thus, every element of $G’’$ is torsion.
Now assume both $G’$ and $G’’$ are torsion, and let $x\in G$. Then $\beta(x)$ is torsion, so there exists $n\gt 0$ such that $e=\beta(x)^n = \beta(x^n)$. Thus, $x^n\in\mathrm{ker}(\beta)=\mathrm{Im}(\alpha)$. Hence, there exists $g\in G’$ such that $\alpha(g)=x^n$. Since $G’$ is torsion, there exists $m\gt 0$ such that $g^m=e$. Then
$$x^{nm} = (x^n)^m = (\alpha(g))^m = \alpha(g^m) = \alpha(e) = e.$$
Thus, $x^{nm}=e$, so $x$ is torsion.
