Show that a nonzero free abelian group has a subgroup of index n for every positive integer n. There is a similar question on here but non of the answers make sense to me.
Show that a nonzero free abelian group has a subgroup of index n for every positive integer n.
2 Answers
The task is equivalent to construct a surjective homomorphism $\to \mathbb{Z}/n$. It is enough to remark that there is a surjective homomorphism $\mathbb{Z} \to \mathbb{Z}/n$ (the canonical projection) and that every nonzero free abelian group $\mathbb{Z}^{\oplus I}$ surjects onto some $\mathbb{Z}$.
Warning: The following seems to require the Axiom of Choice, which with whom I'm very comfortable, btw.
Let $\,A_\omega:=\langle\,a_i\,\rangle_{i\in I}\,$ be the free abelian group group of rank $\,|I|:=\omega\,\,,\,\omega=\text{ any ordinal}$ , and for any $\,n\in\Bbb N\,\,,\,\,C_n:=\langle c_n\,\rangle =$ the cyclic group of order $\,n\,$.
By the universal property of free abelian groups, the set function
$f:\{a_i\}_{i\in I}\to C_n\,\,,\,\,f(a_{i_0}):= c_n\,\,,\,\,f(a_i)=1\,\,,\,i_0\neq i\in I$
extends uniquely to a group homomorphism $\,\phi:A_\omega\to C_n\,$ . (the index $\,i_o\in I\,$ may be chosen at will)
Well, $\,\ker\phi\leq A_\omega\,$ has index $\,n\,$ ...
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0Not for that is AC needed but for defining the function $\,f\,$... – 2012-11-25