Free abelian group $F$ has a subgroup of index $n$? 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.
 A: 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, $\{1\}$). $\Box$
