# Thread: Subgroup of a fraction group

1. ## Subgroup of a fraction group

Let p be a prime and let $\mathbb {Z} (p^ \infty ) = \{ \hat { \frac {a}{b} } \in \frac { \mathbb {Q} }{ \mathbb {Z} } : a,b \in \mathbb {Z} \ , \ b=p^i \ , \ i \geq 0 \}$.

Prove that if there is no upper bound on the orders of elements in a subgroup J of $\mathbb {Z}(p^ \infty)$, then $J = \mathbb {Z}(p^ \infty)$.

Proof so far.

Suppose that J is a subgroup, and suppose that for some elements $j \in J$, $ord(j) = \infty$. So I need to show that some how this element would generate the whole group, how do I do that? Thanks.

Let p be a prime and let $\mathbb {Z} (p^ \infty ) = \{ \hat { \frac {a}{b} } \in \frac { \mathbb {Q} }{ \mathbb {Z} } : a,b \in \mathbb {Z} \ , \ b=p^i \ , \ i \geq 0 \}$.
Prove that if there is no upper bound on the orders of elements in a subgroup J of $\mathbb {Z}(p^ \infty)$, then $J = \mathbb {Z}(p^ \infty)$.
if $\frac{1}{p^n} + \mathbb{Z} \in J,$ for all integers $n \geq 1,$ then $J=\mathbb{Z}_{p^{\infty}},$ and we're done. so let $m=\min \{n \geq 1: \ \frac{1}{p^n} + \mathbb{Z} \notin J \}.$ now suppose $\frac{a}{p^m} + \mathbb{Z} \in J,$ for some $a \in \mathbb{Z}$ with $\gcd(a,p)=1.$ choose $r,s \in \mathbb{Z}$
such that $ra + sp=1.$ then: $\frac{1}{p^m}+\mathbb{Z}=\frac{ra}{p^m} + \frac{s}{p^{m-1}} + \mathbb{Z}\in J.$ contradiction! hence $\frac{a}{p^m} + \mathbb{Z} \notin J,$ for all integers $a$ coprime with $p.$ so: $\forall n \geq m, \forall a \in \mathbb{Z} \ \text{with} \ \gcd(a,p)=1: \ \frac{a}{p^n} + \mathbb{Z} \notin J,$
and therefore $p^{m-1}$ is an upper bound for the orders of elements in $J.$ contradiction! $\Box$