Short Exact Sequences and Torsion

Question

Given a Short Exact Sequence $$\mathbb{Z}_2 \rightarrow A \xrightarrow{\varphi} B \xrightarrow{\psi} \mathbb{Z}_2$$ Where A and B are finite finitely generated abelian groups, I want to show that the odd Torsion part of A is isomorphic to the odd torsion part of B, and the even torsion part of A is isomorphic to the even torsion part of B.

Where I am: We know that A and B are isomorphic to some direct sum, such that $A \cong Tor_2A \oplus Tor_3A \oplus \dots \mathbb{Z}_k$ (where k is prime). Similarly, $B \cong Tor_2B \oplus Tor_3B \oplus \dots \mathbb{Z}_k$ (where k is prime). And this sum for both $A$ and $B$ must be finite, because inifinitely many torsion parts would imply that $A$ and $B$ are not finitely generated or finite. We essentially want to show that $$0 \rightarrow Tor_{odd}A \rightarrow Tor_{odd}B \rightarrow 0$$ By the definition of exact sequences, we know that $\varphi$ is a homomorphism, which would imply that $$ker\varphi \cong 0 \text{ or } \mathbb{Z}_2$$ $$\Rightarrow ker \varphi \cap Tor_{odd} A = 0$$ And $$ Tor_{odd}B \subseteq ker \varphi $$ As if $$\exists \sigma \neq x \in Tor_{odd}B : \psi(x) = 1 \in \mathbb{Z}_2$$ Which should give that $$\mathbb{Z} \psi(x) = 0$ = \psi(2x) = \psi(x + x) = \psi(x) + \psi(x) = 0.$$ And I feel as though I'm almost getting to the implication that odd torsion $\rightarrow 0$, but I'm missing some steps. Any help would be much appreciated!

Answer 1

Your first formula isn't a short exact sequence; I assume you mean to say that

$$0 \to C_2 \rightarrow A \xrightarrow{\varphi} B \xrightarrow{\psi} C_2 \to 0 $$

is a (long) exact sequence, where I've switched to $C_n$ for the cyclic group on $n$ elements.

Proving the odd torsion is the same is easy; the ring $\mathbb{Z}[\frac{1}{2}]$ of all rational numbers whose denominator is a power of $2$ has a flat additive group, so that you get an exact sequence

$$0 \to C_2 \otimes \mathbb{Z}[\tfrac{1}{2}] \rightarrow A \otimes \mathbb{Z}[\tfrac{1}{2}]\xrightarrow{\varphi} B \otimes \mathbb{Z}[\tfrac{1}{2}]\xrightarrow{\psi} C_2 \otimes \mathbb{Z}[\tfrac{1}{2}]\to 0 $$

Since all the groups are torsion groups, this tensor product simply kills off the even torsion, so the result is an exact sequence

$$0 \to 0 \rightarrow A_{\text{odd}} \xrightarrow{\varphi} B_\text{odd} \xrightarrow{\psi} 0 \to 0 $$

Your conjecture about the even torsion is false. Consider the direct sum of the two exact sequences

$$ \begin{matrix} 0 &\to& C_2 &\xrightarrow{\cdot 1}& C_2 &\to& 0 \\ & & 0 &\to& C_2 &\xrightarrow{\cdot 2}& C_4 &\xrightarrow{\cdot 1}& C_2 &\to& 0 \end{matrix} $$

The result is an exact sequence of the desired form, with $A = C_2 \oplus C_2$ and $B = C_4$.