# Short exact sequence terminating at $\mathbb{Z}/2$

Suppose we have the following short exact sequence of abelian groups: $$0 \rightarrow A \xrightarrow{a} B \xrightarrow{b} \mathbb{Z}/2 \rightarrow 0.$$

Under which conditions do we have that $b^{-1}(1)$ is isomorphic, as a set, to $A$?

I suspect this happens when the sequence splits (so then we have $B \cong A \oplus \mathbb{Z}/2$, and the preimage of $1$ under $b$ is just a copy of $A$), but I am having some trouble showing this rigorously. Also, is it true that this sequence always splits?

• I suppose by isomorphism $b^{-1}(1)\cong A$ you mean that of abelian groups. But why should the coset $b^{-1}(1)$ be a group? Commented Nov 14, 2017 at 9:18
• @Phil.Z actually no, it's sufficient for me to show they are isomorphic as sets, I'll add this clarification Commented Nov 14, 2017 at 9:20
• Well, then the answer is always, since $b^{-1}(1)$ is a coset. Commented Nov 14, 2017 at 9:24

If all you want is a bijection between $b^{-1}(1)$ and $A$, then you always have one, as $b^{-1}(1)$ is the coset of $\ker(b)$ not containing zero, and $\ker(b)$ is isomorphic to $A$ by exactness.

• I see, I was overlooking the fact we should view $b^{-1}$ as a coset, so it has the same cardinality as $\ker(b) \cong A$. Thanks! Commented Nov 14, 2017 at 9:31