# order of an element of a group and its coset

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.

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I assumed that e was for $\in$, but I saw the correction when I was done editing. I hope I did not trash the question too much. :) – Asaf Karagila Jun 21 '11 at 16:57
@Asaf: sorry it was my mistake:) – stacy Jun 21 '11 at 17:03

Choose $d$ such that $g+B$ has order $p^d$. Construct $b$ and $g'$ as above. Now just use the definition of order.
@Jack: thank you for your reply. So what I know is that $p^dg=p^db$ and $p^d(g-b)=0$. So the order of $g-b$ is a divisor of $p^d$. I don't understand the reason why I should choose the same $d$ to denote the order of $g+B$ if what I am trying to do is to prove that its order is exactly the same divisor of $p^d$.Maybe I am missing some step. – stacy Jun 21 '11 at 19:02
@Jack: ok, but when I choose $d$ such that $p^d(g+B)=0$ I don't know that $p^d(g-b)=0$ too, since it is what I want to prove. I'm sorry but I don't understand the two implication: $p^d(g+B)=0$ if and only if $p^d(g-b)=0$. $p^d(g+B)=0$ tells me that $p^dg+p^dB=0$.What does it follow ? – stacy Jun 21 '11 at 21:06