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Suppose that we have a free abelian group $F$. How can it be proved that $F$ has a subgroup of index $n$ which $n≥1$?

Honestly, according to the Theorems, I just know that if we take $X$ as a base for $F$, then $$ F= \bigoplus_{\alpha \in X} \mathbb Z_\alpha \ $$ in which for all $ \alpha \in X$; $\mathbb Z_\alpha \ $ is a copy of $ \mathbb Z $. What that subgroup could be? Thanks.

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Looks like you are referring to free abelian groups? –  Grumpy Parsnip May 16 '12 at 15:43
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What about taking the subgroup of $\mathbb{Z}$ generated by $n$ in the first copy, and $\mathbb{Z}$ in the other copies. Would that work? –  M Turgeon May 16 '12 at 15:43
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Whether he means abelian or non-abelian free groups, answering it for free abelian groups is enough, since any non-trivial free group has a non-trivial free abelian group as a quotient. –  Thomas Andrews May 16 '12 at 15:58
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@BabakSorouh He is saying take some $\alpha_0\in X$ and define $H_\alpha$ to be $\mathbb Z_\alpha$ for $\alpha\neq \alpha_0$ and take $H_{\alpha_0}=n\mathbb Z_{\alpha_0}$. Then define $$H = \bigoplus_{\alpha\in X} H_\alpha$$ –  Thomas Andrews May 16 '12 at 16:03
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@Babak: Then say "free abelian". If you say "free group", you mean the absolutely free group. –  Arturo Magidin May 16 '12 at 16:20
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1 Answer

up vote 2 down vote accepted

Theorem. Let $\mathfrak{V}$ be a variety of groups, and let $X$ be a nonempty set. The free $\mathfrak{V}$-group on $X$, $F_{\mathfrak{V}}(X)$ has a subgroup of index $n$ if and only if there is some $|X|$-generated group in $\mathfrak{V}$ with a subgroup of index $n$.

Proof. If $F_{\mathfrak{V}}(X)$ has a subgroup of index $n$, then it witnesses the existence of such a group. Conversely, let $G\in\mathfrak{V}$ be a group, with $\{g_x\}_{x\in X}$ a generating set of $G$, and suppose that $H$ is a subgroup of $G$ of index $n$. The map $f\colon X\to \{g_x\}_{x\in X}$ given by $f(x)=g_x$ induces, by the universal property, a surjective homomorphism $\mathfrak{f}\colon F_{\mathfrak{V}}(X)\to G$. By the isomorphism theorems, $H$ corresponds to a subgroup $\mathcal{H}$ of $F_{\mathfrak{V}}(X)$ that contains $\mathrm{ker}(\mathfrak{f})$, and hence $[F_{\mathfrak{V}}(X):\mathcal{H}] = [G:H] = n$, as claimed. $\Box$

Corollary. If $X\neq\varnothing$, then the free abelian group on $X$ has a subgroup of index $n$ for every positive integer $n$.

Proof. Let $C_n$ be the cyclic group of order $n$. This is $|X|$ generated, and contains a subgroup of index $n$ (namely, $\{1\}$). $\Box$

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Thank you dear Prof. for the answer. May I ask you " if we have this subgroup, then $F$ is an extention of $H$ by $n\mathbb Z_{\alpha_{0}}$" as comments above? Thanks again for the time. –  B. S. May 16 '12 at 16:49
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@Babak: There is no such thing as "this subgroup". There are infinitely many different subgroups of index $n$. Any subgroup of a free abelian group is necessarily free abelian, and a quotient of $F$ by a subgroup of index $n$ is necessarily of order $n$, and so in particular it cannot be torsion free (as it would be if it were $n\mathbb{Z}_{\alpha_0}$. You simply have completely misunderstood the comments above. It seems you just don't understand what is going on at all. –  Arturo Magidin May 16 '12 at 17:18
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