Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let G be an abelian group, $T$ the torsion subgroup of $G$. If $G/T$ is torsion-free, then $T$ and $G/T$ must be disjoint. $G=T \bigoplus G/T$ implies this as well. I don't understand why they are disjoint in the first place.

share|improve this question
2  
This is a mess: $T$ and $G/T$ don't even live in the same group. And aren't you assuming that $G/T$ is finitely generated or something? –  Arturo Magidin Apr 26 '12 at 22:05

2 Answers 2

What you write does not make sense: $T$ "lives" in side $G$, but $G/T$ is a quotient of $G$, not a subgroup. It does not really make any sense to talk about the intersection of $T$ and $G/T$.

What is true is that if $G/T$ is finitely generated, then since $G/T$ is torsion free then it is isomorphic to $\mathbb{Z}^r$. Pick an isomorphism. Then for each of the elements $e_i \in \mathbb{Z}^r$ ($e_i$ has a $1$ in the $i$th coordinate and zeros elsewhere) we can pick an element $g_i\in G$ such that $g_i+T$ corresponds to $e_i$ in $G/T$ under the chosen isomorphism.

Then the $g_i$ generate a subgroup isomorphic to $\mathbb{Z}^r$: if $a_1g_1+\cdots+a_rg_r=0$, then projecting onto $G/T$ we get $a_1e_1+\cdots+a_re_r=0$, hence $a_i=0$ for all $i$; so $\langle g_1,\ldots,g_r\rangle = \langle g_1\rangle\oplus\langle g_2\rangle\oplus\cdots\oplus \langle g_r\rangle$.

And now we have the important point: $\langle g_1,\ldots,g_r\rangle\cap T = \{0\}$. For if $a_1g_1+\cdots+a_rg_r\in T$, then there exists $n\gt 0$ such that $n(a_1g_1+\cdots+a_rg_r) = 0$. This gives $$na_1g_1+\cdots+na_rg_r=0,$$ hence $na_i=0$ for all $i$, so $a_i=0$ for all $i$. That is, the intersection is contained in $\{0\}$.

Since $T+\langle g_1,\ldots,g_r\rangle = G$ (e.g., by the isomorphism theorems), and the intersection is trivial, we get $$G \cong T\oplus \langle g_1,\ldots,g_r\rangle \cong T\oplus \mathbb{Z}^r\cong T\oplus (G/T).$$

share|improve this answer

What is not true. If $G=\prod_{p}\mathbb{F}_p=\mathbb{F}_2\times\mathbb{F}_3\times\cdots$, then $G/T$ can not inject to $G$. In particular, $G$ is not isomorphic to $T\oplus G/T$.

Proof It is clear that $\bigcap_ppG=0$. It suffices to show $\cap_pp(G/T)\neq 0$. The image of $x=(1,1,\ldots)$ is in $p(G/T)$ for every prime $p$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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