1
$\begingroup$

Here is the problem:

Let $G=\langle x_0,x_1,x_2,\ldots\ |px_0=0,x_0=p^nx_n, \text{all } n\geq1\rangle$. Prove that $G/\langle x_0\rangle$ is a direct sum of cyclic groups and is reduced.

The first part is easy because if we put the relation $x_0=0$ to other relations in $G$; $G$ would be a direct sum of cyclic groups. For another part, I feel that the first part is usefull, but I don't know how to link these together. The following ideas just came to me:

  • if I assume it is not reduced; $dG$ would be a proper subgroup and so $G/dG$ is reduced.

or

  • To show that $\{0\}$ is the only divisible subgroup of $G$.

Thanks for your time.

  • 0
    Exactly. I wanted to do that.2012-10-12

2 Answers 2

1

Consider $p^\infty G := \bigcap_{n=1}^\infty p^n G$. For each $n$, $p^n (A/B) = (p^nA+B)/B$, but for $A=G$, $B=\langle x_0 \rangle$, $p^n(A/B) \leq \langle \bar x_n, \dots \rangle$, so $p^n A \leq \langle x_0, x_n,x_{n+1}, \ldots\rangle$, so $p^\infty A = \langle x_0 \rangle$. Hence every divisible subgroup of $G$ is contained within $p^\infty G = \langle x_0 \rangle$ of order $p$, so $dG=0$.

When working with abelian $p$-groups, you'll mostly be concerned with these $p^n A$ and $A[p^n]$ subgroups, and their transfinite counter parts.

$p^{\infty+1}G = p(p^\infty(G)) = 0$, but $dG \leq p^{\infty+\infty}(G) = p^\infty(p^\infty(G)) \leq p^{\infty+1}(G)$.

A fun exercise is to find an abelian $p$-group with $p^{2\infty}(G) \neq 0$, but $p^{3\infty}(G) = 0$.

  • 0
    That is very interesting working with that infinity powers. I am working on one of the Rotman’s book and have not seen these magic powers. Thank you for sharing me your knowledge. You pointed that, since every divisible subgroup is contained in $p^{\infty}G\cong \mathbb Z_p$ then $dG=0$. Is this fact arisen from the point that **finite groups are not divisible** ? However, I am thinking to find out why did you say "**Hence every...of order $p$**".2012-10-13
2

Suppose $\,A\leq G/\langle\,x_0\,\rangle\,$ is a divisible subgroup and let $\,a\in A\,$ . Then

$\forall\,n\in\Bbb N\,\,\exists\,a_n\in A\,\,s.t.\, a= na_n$

Now, we can write $\,a=w(x_i)+\langle\,x_0\,\rangle\,$ ,where $\,w(x_i)\,$ is a word in a finite number of generators $\,x_i\,$ of $\,G\,$.

Well, let now $\,m:=\max\,\{i\;;\;x_i\,\,\text{appears in the word}\,\,w\}$ , and take now $\,n:=p^m\,$ above...can you take it from here?