2
$\begingroup$

I find it hard to understand a part of the proof of the existence of any basic subgroup in every abelian torsion group.I'm going to write you the information I think useful.

Let $G$ an abelian torsion group. Let $B=\langle X\rangle$ where $X$ is a maximal pure independent subset. If we prove that $G/B$ is divisible it follows that $B$ is basic and so our thesis.

Suppose by contradiction that $G / B$ is not divisible, hence it has a nontrivial pure cyclic subgroup $\langle g+B\rangle$. By purity of $B$ we have that if $ p^dg$ belongs to $B$, then $p^dg$ belongs to $p^dB$. Hence $p^dg=p^db$ and $p^d(g-b)=0$ where $b\in B$.

Since $(g-b)+B=g+B$ it follows that g'=g-b and g'+B have the same order.

Why is the latter true? How can I prove that g'=g-b and g'+B have the same order? I hope I gave you all relevant information.

Can anyone help me? Thanks.

  • 0
    @Asaf: sorry it was my mistake:)2011-06-21

1 Answers 1

2

Choose $d$ such that $g+B$ has order $p^d$. Construct $b$ and g' as above. Now just use the definition of order.

Let me know if you need a stronger hint (or view the answer's source), but I don't see any obstacle.

  • 0
    Exactly! No problem.2011-06-22