Proof of Prüfer Group Non-Existence

[charlamov]

proove that
< x0 , x1 , . . . | [xi , xj ] = 1, i, j, ∈ N_0 ; x0^p = 1 ; (xi) ^ (p ^ i) = x0 , i ∈ N > is not presentation of Prüfer group

 

[DonAntonio]

proove that
< x0 , x1 , . . . | [xi , xj ] = 1, i, j, ∈ N_0 ; x0^p = 1 ; (xi) ^ (p ^ i) = x0 , i ∈ N > is not presentation of Prüfer group

Please do learn quickly how to type in LaTeX in this site: https://www.physicsforums.com/showthread.php?t=546968

I'll try to edit your post (and, perhaps, address it):

Prove (please, of course), that $$\langle x_0,x_1,...\,\,|\,\,[x_i,x_j]=1\,,\,i,j\in\mathbb{N}\,,\,x_0^p=1\,,\,x_i^{p^i}=x_0\,,\,i\in\mathbb{N}\rangle$$ is not a presentation of the Prüfer group.

Now, the Prüfer group must fulfill the conditions $\,x^p_i=x_{i-1}\,,\,i=1,2,3,...\,$, but by your definition we'd have $$x_1^p=x_0\,,\,x_2^{p^2}=x_0=x_1^p\Longrightarrow$$and I can't see how we can deduce from this that $\,x_2^p=x_1\,$ , as we're not sure we can take $p-th$ roots...

Another possible approach: to show that an epimorphism from an infinitely generated abelian free sugroup to one of the groups is not the same for the other one...

DonAntonio