Proper subgroups of fraction group

**tttcomrader** · November 29th 2008, 08:51 PM

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 the only proper subgroups of $\mathbb {Z} (p^ \infty )$ are the finite cylic groups $J_k = < \hat { \frac {1}{p^k} } >$

Proof so far.

Suppose to the contrary that H is a subgroup of $\mathbb {Z} (p^ \infty )$ such that $H \neq \mathbb {Z} (p^ \infty )$ and $H \neq < \hat { \frac {1}{p^k} } > \ \ \ \forall k$ . Now, suppose that $H = < \frac {a}{b}>$ , then $\frac {a}{b} \neq \frac {1}{p^k}$ . But how would I get a contradiction? Thanks.

**NonCommAlg** · November 29th 2008, 09:39 PM

Originally Posted by tttcomrader

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 the only proper subgroups of $\mathbb {Z} (p^ \infty )$ are the finite cylic groups $J_k = < \hat { \frac {1}{p^k} } >$

let $J$ be a proper subgroup of $\mathbb{Z}_{p^{\infty}}.$ then there exists an integer $n \geq 1$ such that $\frac{1}{p^n} + \mathbb{Z} \notin J.$ let $m=\min \{n \geq 1: \ \frac{1}{p^n} + \mathbb{Z} \}.$ then $\frac{a}{p^n} + \mathbb{Z} \in J,$ for all $a \in \mathbb{Z}$ and $0 \leq n \leq m-1.$

in here i showed that $\frac{a}{p^n} + \mathbb{Z} \notin J,$ for all integers $a$ coprime with $p$ and for all $n \geq m.$ therefore: $J=\{\frac{a}{p^n}+\mathbb{Z}: \ \ a \in \mathbb{Z}, \ 0 \leq n \leq m-1 \} = <\frac{1}{p^{m-1}} + \mathbb{Z} >. \ \Box$

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