exact sequence in group, $0 \to \ker f \to A \to B \to 0$ and subgroup

Question

Let $f:A \to B$ be a group surjective homomorphism, There is a exact sequence in group, $0 \to \ker f \to A \to B \to 0$.

Suppose $\ker f$ has no torsion element, and $A$ has subgroup isomorphic to Klein four group $\mathbb{Z/2Z\times Z/2Z}$.

Then, why $B$ has subgroup isomorphic to $\mathbb{Z/2Z\times Z/2Z}$ ?

My though: If this exact sequence spilts, then every torsion part of $A$ is to $B$, so I can say the result. But in general, I think this exact sequence does not spilt, so I'm stuck.

$A/ \ker f$ is isomorphic to $B$, from here, can I say something・・　？

Answer 1

More generally let $H\subseteq A$ be a torsion (periodic) subgroup and let $f:A\to B$ be a group homomorphism such that $\ker f$ is torsion free. Then $H$ and $f(H)$ are isomorphic.

Indeed, let $x\in H$, $x\neq e$. Then $f(x)\neq e$ because otherwise $x$ would belong to $\ker f$ which is torsion free while $x$ is of finite order. It follows that $f_{|H}$ is injective (it has trivial kernel) and thus the restriction $f_{|H}:H\to f(H)$ is an isomorphism.

Answer 2

Let $H$ be the subgroup of $A$ isomorphic to $C_2 \times C_2$, and let $K = \ker f$ (so $A/K \cong B$).

Then $H \cap K = 1$, so $HK/K \cong H/(H \cap K) \cong H$.