Let $H$ be a group, if its abelianization is $\mathbb{Z}_2$ does this mean that $H$ has torsion?

As is mentioned in the title, I have some group $H$ and I know that its abelianization is $\mathbb{Z}_2$. Does this imply that $H$ has torsion?

Edit: Since people want more context, here's some context. Basically I'm looking at the fundamental group of the Klein bottle and I want to show that it can't split as $\pi_1(K) \cong \mathbb{Z} \oplus H$ for any group $H$. I know if I abelianize then I end up getting $\mathbb{Z} \oplus \mathbb{Z}_2$. So if the abelianization splits over direct sums (which I'm not sure about) then if somehow I could say that $H$ must be going to $\mathbb{Z}_2$ through this abelianization map and $H$ had to have torsion I'd have a contradiction to the klein bottle being a manifold and thus having torsion free fundamental group. I think there are a number of holes in this argument though, but I'm still interested in the particular question I asked above.

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