2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Every abelian group is the direct sum of its Sylow ssubgroups, and every morphism of groups preserves that decomposition. This implies that your exact sequence splits in one dshort exact sequence for the even parts and one for the odd part. $\endgroup$ – Mariano Suárez-Álvarez Mar 28 '17 at 2:00
0
$\begingroup$

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 $B = C_4$.

$\endgroup$
  • $\begingroup$ Hi! I'm unfamiliar with the term "flat additive group." Also, this works if I replace the $C_2$s with $\mathbb{Z}_2$? $\endgroup$ – Ravmcgav Mar 28 '17 at 10:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.