Show that Q, the group of rational numbers under addition, has no proper subgroups of finite index, but Z has.

Printable View

- Oct 26th 2008, 05:27 PMmandy123show has no proper subgroups of finite index
Show that Q, the group of rational numbers under addition, has no proper subgroups of finite index, but Z has.

- Oct 26th 2008, 07:36 PMNonCommAlg
let N be a subgroup of $\displaystyle \mathbb{Q}$ with $\displaystyle [\mathbb{Q}:N]=n.$ since $\displaystyle (\mathbb{Q},+)$ is abelian, N is normal and hence $\displaystyle \mathbb{Q}/N$ is a group of order $\displaystyle n.$ so: $\displaystyle \mathbb{Q}=n\mathbb{Q} \subseteq N \subseteq \mathbb{Q}.$ thus: $\displaystyle N=\mathbb{Q}. \ \ \Box$

is this really not in your lecture notes that every nonzero subgroup of $\displaystyle \mathbb{Z}$ has finite index?!!! (Dull)