# Thread: Subgroup of a fraction group

1. ## Subgroup of a fraction group

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

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

Proof so far.

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