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I have the following problem$^*$:

Prove that the group $G$ having generators and relations respectively $X=\{x_0,x_1,x_2,\ldots\} \\\{px_0=0,x_0=p^nx_n, \text{all } n\geq1\}$ is an infinite $p$-primary group with $\bigcap_{n=1}^\infty p^nG\neq0$.

What I have done:

Since the set of generators are infinite so is the group. For another claim, I suppose $F$ to be free abelian group on $X$ and let the subgroup of it, say $R$ generated by the relations. Now $a_n=x_n+R\in G$ and $pa_0=0, p^2a_1=pa_0=0,p^3a_2=p^2x_0=0,\ldots$ and so $p^{n+1}a_n=0$. I conclude the group is $p$-primary. For the last I see that $x_0$ is an element in that intersection.

Is my approach right? Did I conclude correctly that the above intersection has $x_0$ in itself or the intersection can have other elements than $x_0$? Thanks.

$(*)$ : An Introduction to the Group Theory by J.J.Rotman pp. 317 problem 10.5

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Your justification for G being infinite is, I believe, a little lacking: after all, we've relations connecting every generator to one single generator so it could possibly be that all the generators would "collapse" into one single one with finite order and thus the group would be finite. It is not the case but something must be said, though.

I'd go like this: since

$\forall\,n\in\Bbb N\;\;,\;\;x_n=p^{-n}x_0\Longrightarrow G\,\,\text{ is abelian}\,$

Also, since $\,px_0=0\,$ , we get from the above that $\,p^{n+1}x_n=px_0=0\,$ , so the group is an abelian $\,p-\,$ group .

Also, as you correctly pointed out, $\,x_0\in p^nG\,\,,\,\forall\,n\in\Bbb N\,$ .

Finally: $\,x_0=x_1\Longrightarrow x_1=x_0=px_1\Longrightarrow\,$ the order of $\,x_1\,$ is a divisor of $\,p-1\,$ and thus is not a power of $\,p\,$, contradicting the above.

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    @BabakSorouh: correct, it is a different group. It's "ulm invariant", rank(G[p^2]/G[p]) = infinity is different than for the Prüfer group, with rank 1. Here G[n] = { g : ng = 0 }. This is a group where an element $x_0$ has infinite height but is not divisible (so for any $n$, $x_0$ can be divided by $p^n$, and yet, $x_0$ cannot be repeatedly divided by $p$, since once you divide it by $p$ to get one of the $p^{n-1}x_n$, you have only a limited number of divisions by $p$ left).2012-10-10