Free abelian group

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In abstract algebra, a free abelian group is an abelian group that has a "basis" in the sense that every element of the group can be written in one and only one way as a finite linear combination of elements of the basis, with integer coefficients. Unlike vector spaces, not all abelian groups have a basis, hence the special name for those that do. A typical example of a free abelian group is the direct sum Z ⊕ Z of two copies of the infinite cyclic group Z; a basis is {(1,0),(0,1)}. The trivial abelian group {0} is also considered to be free abelian, with basis the empty set.

A point on terminology: a free abelian group is not the same as a free group that is abelian; a free abelian group is not necessarily a free group. In fact the only free groups that are abelian are those of rank 0 (the trivial group) and rank 1 (the infinite cyclic group).

If F is a free abelian group with basis B, then we have the following universal property: for every arbitrary function f from B to some abelian group A, there exists a unique group homomorphism from F to A which extends f. This universal property can also be used to define free abelian groups.

For every set B, there exists a free abelian group with basis B, and all such free abelian groups having B as basis are isomorphic. One example may be constructed as the abelian group of functions on B, taking integer values all but finitely many of which are zero. This is the direct sum of copies of Z, one copy for each element of B. Formal sums of elements of a given set B are nothing but the elements of the free abelian group with basis B.

Every finitely generated free abelian group is therefore isomorphic to Zⁿ for some natural number n called the rank of the free abelian group. In general, a free abelian group F has many different bases, but all bases have the same cardinality, and this cardinality is called the rank of F. This rank of free abelian groups can be used to define the rank of all other abelian groups: see rank of an abelian group. The relationships between different bases can be interesting; for example, the different possibilities for choosing a basis for the free abelian group of rank two is reviewed in the article on the fundamental pair of periods.

Given any abelian group A, there always exists a free abelian group F and a surjective group homomorphism from F to A. This follows from the universal property mentioned above.

Importantly, every subgroup of a free abelian group is free abelian (proof in the end of the article). As a consequence, to every abelian group A there exists a short exact sequence

0 → G → F → A → 0

with F and G being free abelian (which means that A is isomorphic to the factor group F/G). This is called a free resolution of A. Furthermore, the free abelian groups are precisely the projective objects in the category of abelian groups.

All free abelian groups are torsion-free, and all finitely generated torsion-free abelian groups are free abelian. (The same applies to flatness, since an abelian group is torsion-free if and only if it is flat.) The additive group of rational numbers Q is a (not finitely generated) torsion-free group that's not free abelian. The reason: Q is divisible but non-zero free abelian groups are never divisible.

Free abelian groups are a special case of free modules, as abelian groups are nothing but modules over the ring Z.

It can be surprisingly difficult to determine whether a concretely given group is free abelian. Consider for instance the Baer-Specker group Z^N, the direct product of countably many copies of Z. Reinhold Baer proved in 1937 that this group is not free abelian; Specker proved in 1950 that every countable subgroup of Z^N is free abelian.

[edit] Subgroups of free abelian groups are free

This is related to the Nielsen-Schreier theorem that a subgroup of a free group is free.

Theorem: Let $F$ be a free abelian group generated by the set $X=\{x_k\,|\,k\in I\}$ and let $G\subset F$ be a subgroup. Then $G$ is free.

Proof: This proof is an application of Zorn's lemma and can be found in Serge Lang's Algebra, Appendix 2 §2. If $G = {0}$ , the statement holds, so we can assume that $G$ is non-trivial.

If $G = {0}$ , the statement holds, so we can assume that $G$ is non-trivial. First we shall prove this for finite $X$ by induction. When $| X | = 1$ , $G$ is isomorphic to $\Z$ (being non-trivial) and clearly free. Assume that if a group is generated by a set of size $\le k$ , then every subgroup of it is free. Let $X=\{x_1,x_2,\dots,x_k,x_{k+1}\}$ , $F$ the free group generated by $X$ and $G\subset F$ a subgroup. Let $\operatorname{pr}\colon G\to F$ be the projection $\operatorname{pr}(a_1x_1+\dots+a_{k+1}x_{k+1})=a_1x_1.$ If $\operatorname{Rng}(\operatorname{pr})=\{0\}$ , then $G$ is a subset of $\langle x_2,\dots,x_{k+1}\rangle$ and free by the induction hypothesis. Thus we can assume that the range is non-trivial. Let $m > 0$ be the least such that $mx_1\in \operatorname{Rng}(\operatorname{pr})$ and choose some $x$ such that $\operatorname{pr} x=mx_1$ . It is standard to verify that $x\notin\operatorname{Ker}(\operatorname{pr})$ and if $y\in G$ , then $y = n x + k$ , where $k\in \operatorname{Ker}(\operatorname{pr})$ and $n\in\Z$ . Hence $G=\operatorname{Ker}(\operatorname{pr}) \oplus \langle x\rangle$ . By the induction hypothesis $\operatorname{Ker}(\operatorname{pr})$ and $\langle x\rangle$ are free: first is isomorphic to a subgroup of $\langle x_2,\dots x_{k+1}\rangle$ and the second to $\Z$ .

Assume now that $X=\{x_i\mid i\in I\}$ is arbitrary. For each subset $J$ of $I$ let $F J$ be the free group generated by $\{x_i\mid i\in J\}$ , thus $F_J\subset F$ is a free subgroup and denote $G_J=F_J\cap G$ .

Now set

$S=\{(G_J,w)\mid G_J {\rm\; is\; a\; non-trivial\; free\; group\;and\; }w{\rm\; is\; a\; basis\; of\;}G_J\}.$

Formally $w$ is an injective (one-to-one) map

$w:J'\to G_J$

such that $w [J']$ generates $G J$ .

Clearly $S$ is non-empty: Let us have an element $x$ in $G$ . Then $x=a_1x_{i_1}+\cdots+a_nx_{i_n}$ and thus the free group generated by $S=\{x_{i_1},\dots, x_{i_n}\}$ contains $x$ and the intersection $G\cap F_J$ is a non-trivial subgroup of a finitely generated free abelian group and thus free by the induction above.

If $(G_J,w),(G_K,u)\in S$ , define order $(G_J,w)\le(G_K,u)$ if and only if $J\subset K$ and the basis $u$ is an extension of $w$ ; formally if $w:J'\to G_J$ and $u:K'\to G_K$ , then $J'\subset K'$ and $u\restriction w=w$ .

If $(G_{J_r},w_r)_{r\in L}$ is a $\le$ -chain ( $L$ is some linear order) of elements of $S$ , then obiously

$(\bigcup_{r\in L}G_{J_r},\bigcup_{r\in L}w_r)\in S$ ,

so we can apply Zorn's lemma and conclude that there exists a maximal $(G J, w)$ . Since $G I = G$ , it is enough to prove now that $J = I$ . Assume on contrary that there is $k\in I\setminus J$ .

Put $K=J\cup\{k\}$ . If $G_K=F_K\cap G=G_J,$ then it means that $(G_J,w)\le (G_K,w)$ , but they are not equal, so $(G K, w)$ is bigger, which contradicts maximality of $(G J, w)$ . Otherwise there is an element $nx_k + y\in G_K$ such that $n\in\Z\setminus\{0\}$ and $y\in G_J\subset F_J$ .

The set of $n\in\Z$ for which there exists $y\in G_J$ such that $nx_k+y\in G$ forms a subgroup of $\Z$ . Let $n 0$ be a generator of this group and let $w_k=n_0x_k+y\in G,$ with $y\in F_J$ . Now if $z\in G_K$ , then for some $m\in Z$ $z = z - m w k + m w k$ , where $z-mw_k\in G_J$ .

On the other hand clearly $w_k\Z\cap G_J=\{0\}$ , so $w'=w\cup \{\langle k,w_k\rangle\}$ is a basis of $G K$ , so $(G_K,w')\ge (G_J,w)$ contradicting the maximality again. $\square$