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Let $A$ be a free abelian group of finite rank and $B$ be a subgroup of $A$ such that $A=B+pA$ for some prime number $p$. Then how to prove $B$ is a subgroup of finite index in $A$? And if $A=B+pA$ holds for any prime number number $p$, then $A=B$?

I tried to use The Second Isomorphism Theorem for groups, since any subgroup of an abelian group is normal, so we can get $pA/(pA\cap B) \cong\ (B+pA)/B=A/B$, since $A=B+pA$, so the next step will be show $pA/(pA\cap B)$ is finite, then I got stuck. I also tried to use $A/pA=(B+pA)/pA \cong\ B/(pA\cap B)$, and stuck again.

I guess we might need to connect $A/B$ with $A/pA$, since the later the finite, and I don't know how to use the condition that A is with finite rank, any suggestions?

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    It seems like the theorem on [elementary divisors](http://math.stanford.edu/~conrad/210APage/handouts/strthm.pdf) would blow this away, but that probably isn't available to you.2011-08-05
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    As Dylan mentioned, it's hard to give a hint/answer without knowing what you're comfortable with. Note that it is enough to show every element of $A/B$ has finite order. Since $A=B+pA$, every element in $A/B$ has a p-th root. Now you can proceed in several ways, but basically an element of infinite order would give you an infinite ascending chain of subgroups in $A/B$, which is impossible.2011-08-05
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    For the second question: once we know $A/B$ is finite, and every element has a p-th root, it follows that $p$ does not divide $|A/B|$; doing this for every prime shows $|A/B|=1$.2011-08-05
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    hey,Steve,what do you mean by every element in $A/B$ has a p-th root? Actually I'm not sure how to represent an element in $A/B$?2011-08-05
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    Is this correct? Following @Steve D: Let $c$ be an element of infinite order in $C:=A/B$, for $n=1,2,...$ let $c_n\in C$ satisfy $p^ nc_n=c$, let $C_n$ be the subgroup generated by $c_n$. There is an $n$ such that $c_{n+1}\in C_n$, contradiction.2011-08-05

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