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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.

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2 Answers 2

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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 $\mathbb{Z}$.

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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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  • $\begingroup$ The argument is of courdse not restricted to finitely generated free-abelian groups ... $\endgroup$ Nov 25, 2012 at 18:45
  • $\begingroup$ Of course, my bad. Thanks and I shall remove that restriction now $\endgroup$
    – DonAntonio
    Nov 25, 2012 at 18:46
  • $\begingroup$ I think you don't need AC. That the set of subgroups of index $n$ is non-empty should follow from the fact that every element of $I$ gives rise to such a subgroup. $\endgroup$ Nov 25, 2012 at 18:58
  • $\begingroup$ Not for that is AC needed but for defining the function $\,f\,$... $\endgroup$
    – DonAntonio
    Nov 25, 2012 at 22:30

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