Subgroup of Index ##n## for every ##n \in \Bbb{N}##.

[Bashyboy]

Homework Statement

A nonzero free abelian group has a subgroup of index $n$ for every positive integer $n$

Homework Equations

The Attempt at a Solution

If $F$ is a nonzero free abelian group, then $F$ is isomorphic to the direct sum $G= \sum_{i \in I} \Bbb{Z}$, where $I \neq \emptyset$. Since $I$ is not empty, it contains at least one element $k$. Now define $H_k = n \Bbb{Z}$, and define $H_i = \Bbb{Z}$ for all $i \neq k$. Then clearly $H := \sum_{i \in I} \Bbb{Z}$ is a subgroup of $G$. Since $G$ is abelian, $H$ is trivially normal and therefore $G/H \simeq \sum_{i \in I} \Bbb{Z}/H_i \simeq \Bbb{Z}_n$, which has order cardinality $n$. Hence, $H$ is a subgroup of index $n$, and since $G$ is isomorphic to $F$, it too must have a subgroup of index $n$.

How does this sound?

 

[fresh_42]

Sounds good. I don't know why you mentioned normality, but I can't see any wrongs.