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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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up vote 2 down vote accepted

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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So, we can use what you pointed finally to show that all $a_n$ are distinct and then $G$ would be infinite? –  B. S. Oct 10 '12 at 12:41
    
Well, yes. What I wrote is "the hint" to continue inductively. –  DonAntonio Oct 10 '12 at 12:43
    
Thanks Don for every words. –  B. S. Oct 10 '12 at 12:45
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For the finally, it might be easier to mod out by $x_0$, then you have the presentation of the group $$\bigoplus_{n=1}^\infty \mathbb{Z}/p^n\mathbb{Z}$$. Since a quotient of your group is infinite, the group itself is infinite (and those $x_n$ are distinct from each other for $n>0$, even mod $x_0$). –  Jack Schmidt Oct 10 '12 at 13:12
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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). –  Jack Schmidt Oct 10 '12 at 16:07
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