1
$\begingroup$

$$0\rightarrow G_{1}\rightarrow G_{2}\rightarrow G_{3}\rightarrow 0$$ is a short exact sequence of finitely generated abelian groups. We call $\bar{G_{i}}$ the quotient of $G_{i}$ by its torsion subgroup.

I want to show that $$0\rightarrow \bar{G_{1}}\rightarrow \bar{G_{2}}\rightarrow \bar{G_{3}}\rightarrow 0$$ is not exact. I am looking for example with $\mathbb{Z}/n\mathbb{Z}$ or something else... Can you help me ? Thank you.

$\endgroup$

2 Answers 2

1
$\begingroup$

A simple example is to take a sequence $$0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/n\to 0$$ The sequence becomes $$0\to \mathbb{Z}\stackrel{\times n}{\to} \mathbb{Z}\to 0$$ This is not an exact sequence.

$\endgroup$
1
  • $\begingroup$ Cause you don't write the $0$ for $(\mathbb{Z}/n\mathbb{Z})/(\mathbb{Z}/n\mathbb{Z})$ ? $\endgroup$ Commented Jan 2, 2015 at 15:03
0
$\begingroup$

Suppose your sequence is always exact. Then take $G_2$ any torsionfree group and $G_1$ any subgroup of $G_1$. Then also $G_1$ is torsionfree and exactness would imply that $G_2/G_1$ is torsionfree.

But for every abelian group $G$ there exists a surjection $F\to G$, with $F$ a free group (in particular torsionfree). So every group would be torsionfree.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .