# show has no proper subgroups of finite index

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• October 26th 2008, 05:27 PM
mandy123
show 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.
• October 26th 2008, 07:36 PM
NonCommAlg
Quote:

Originally Posted by mandy123
Show that Q, the group of rational numbers under addition, has no proper subgroups of finite index, but Z has.

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

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