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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
up vote 2 down vote accepted

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.

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@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
@Stacy: Ah, this is a little confusing. In the proof you gave above, "d" was never fixed. The proof just said that if such and such (involving d) then such and such. It worked generally for lots of d (that need not be exponent of p in the order of anything). Well, we need to choose a good d for it to work on. Once you choose that d, everything works out nicely. – Jack Schmidt Jun 21 '11 at 19:23
@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
@Stacy: p^d*(g+B) = 0 means exactly that (p^dg) + B = 0 + B, which means exactly that p^dg is in B. – Jack Schmidt Jun 22 '11 at 0:54
@Jack: And now I can use the purity of B as in the proof. Thank you – stacy Jun 22 '11 at 7:33

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