- Free abelian group
In

abstract algebra , a**free abelian group**is anabelian 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 finitelinear combination of elements of the basis, withinteger coefficients. Unlikevector space s, not all abelian groups have a basis, hence the special name for those that do. (For instance, any group having periodic elements is not a free abelian group because any element can be expressed in an infinite number of ways simply by putting in an arbitrary number of cycles constructed from a periodic element.)A typical example of a free abelian group is the direct sum**Z**⊕**Z**of two copies of the infinitecyclic group **Z**; a basis is {(1,0),(0,1)}. The trivial abelian group {0} is also considered to be free abelian, with basis theempty 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 having an empty basis (rank 0, giving thetrivial group ) or having just 1 element in the basis (rank 1, giving theinfinite cyclic group ).Other abelian groups are not "free groups" because in free groups "ab" must be different from "ba" if "a" and "b" are different elements of the basis.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 uniquegroup 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

**Z**^{"n"}for somenatural 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 samecardinality , 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: seerank 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 thefundamental 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 ashort exact sequence :0 → "G" → "F" → "A" → 0with "F" and "G" being free abelian (which means that "A" is isomorphic to thefactor group "F"/"G"). This is called a**free resolution**of "A". Furthermore, the free abelian groups are precisely the projective objects in thecategory of abelian groups . [*Griffith, p.18*]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 number s**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 module s, 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}, thedirect 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.**ubgroups of free abelian groups are free**This is related to the

Nielsen-Schreier theorem that a subgroup of afree group is free.Theorem : Let $F$ be a free abelian group generated by the set $X=\{x\_k,|,kin\; I\}$ and let $Gsubset\; F$ be a subgroup. Then$G$ is free.Proof : This proof is an application ofZorn's lemma and can be found inSerge Lang 's Algebra, Appendix 2 §2.If $G=\{0\}$, the statement holds, so we can assume that $G$ is nontrivial.First we shall prove this for finite $X$ by induction. When $|X|=1$, $G$is isomorphic to $$ (being nontrivial) and clearly free. Assume that if a group is generated by a set of size $le\; k$, then every subgroup of it is free. Let $X=\{x\_1,x\_2,dots,x\_k,x\_\{k+1\}\}$, $F$ the free group generatedby $X$ and $Gsubset\; F$ a subgroup. Let $operatorname\{pr\}colon\; G\; o\; F$ be the projection$operatorname\{pr\}(a\_1x\_1+dots+a\_\{k+1\}x\_\{k+1\})=a\_1x\_1.$ If $operatorname\{Rng\}(operatorname\{pr\})=\{0\}$, then $G$ is a subset of $langle\; x\_2,dots,x\_\{k+1\}\; angle$ and freeby the induction hypothesis. Thus we can assume that the range is nontrivial.Let $m>0$ be the least such that$mx\_1in\; operatorname\{Rng\}(operatorname\{pr\})$ and choose some $x$ such that $operatorname\{pr\}\; x=mx\_1$. Itis standard to verify that$x\; otinoperatorname\{Ker\}(operatorname\{pr\})$ and if $yin\; G$, then $y=nx+k$, where $kin\; operatorname\{Ker\}(operatorname\{pr\})$ and $nin$.Hence $G=operatorname\{Ker\}(operatorname\{pr\})\; oplus\; langle\; x\; angle$.By the induction hypothesis $operatorname\{Ker\}(operatorname\{pr\})$ and $langle\; x\; angle$ are free:first is isomorphic to a subgroup of $langle\; x\_2,dots\; x\_\{k+1\}\; angle$ and the second to $$.

Assume now that $X=\{x\_imid\; iin\; I\}$ is arbitrary.For each subset $J$ of $I$ let $F\_J$ be the free group generated by $\{x\_imid\; iin\; J\}$,thus $F\_Jsubset\; F$is a free subgroup and denote $G\_J=F\_Jcap\; G$.

Now set

$S=\{(G\_J,w)mid\; G\_J\; \{\; m;\; is;\; a;\; nontrivial;\; free;\; group;and;\; \}w\{\; m;\; is;\; a;\; basis;\; of;\}G\_J\}.$

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

$w:J\text{'}\; o\; G\_J$

such that $w\; [J\text{'}]$ generates $G\_J$.

Clearly $S$ is nonempty: Let us have an element $x$ in $G$.Then $x=a\_1x\_\{i\_1\}+cdots+a\_nx\_\{i\_n\}$ and thus the free groupgenerated by $S=\{x\_\{i\_1\},dots,\; x\_\{i\_n\}\}$ contains $x$ and theintersection $Gcap\; F\_J$ is a nontrivial 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 $Jsubset\; K$ and the basis $u$ is an extension of $w$; formallyif $w:J\text{'}\; o\; G\_J$ and $u:K\text{'}\; o\; G\_K$, then $J\text{'}subset\; K\text{'}$ and $u\; estriction\; w=w$.

If $(G\_\{J\_r\},w\_r)\_\{rin\; L\}$is a $le$-chain ($L$ is some linear order) of elementsof $S$, then obiously

$(igcup\_\{rin\; L\}G\_\{J\_r\},igcup\_\{rin\; L\}w\_r)in\; S$,

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 $kin\; Isetminus\; J$.Put $K=Jcup\{k\}$. If $G\_K=F\_Kcap\; G=G\_J,$ 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 $nx\_k\; +\; yin\; G\_K$such that $ninsetminus\{0\}$ and $yin\; G\_Jsubset\; F\_J$.

The set of $nin$ for which there exists $yin\; G\_J$ such that$nx\_k+yin\; G$ forms a subgroup of $$. Let $n\_0$ bea generator of this group and let $w\_k=n\_0x\_k+yin\; G,$with $yin\; F\_J$. Now if $zin\; G\_K$, then for some $min\; Z$$z=z-mw\_k+mw\_k$, where $z-mw\_kin\; G\_J$.

On the other handclearly $w\_kcap\; G\_J=\{0\}$, so $w\text{'}=wcup\; \{langle\; k,w\_k\; angle\}$is a basis of $G\_K$, so $(G\_K,w\text{'})ge\; (G\_J,w)$ contradicting the maximality again. $square$

**ee also***

Finitely generated abelian group **References***

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