Free abelian group

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. Hence, free abelian groups over a basis B are also known as formal sums over B. Informally, free abelian groups or formal sums may also be seen as signed multisets with elements in B.

Free abelian groups have very nice properties which make them similar to vector spaces and allow a general abelian group to be understood as a quotient of a free abelian group by "relations". Every free abelian group has a rank defined as the cardinality of a basis. The rank determines the group up to isomorphism, and the elements of such a group can be written as finite formal sums of the basis elements. Every subgroup of a free abelian group is itself free abelian, which is important for the description of a general abelian group as a cokernel of a homomorphism between free abelian groups.

Example

For example, let G be the group that is the direct sum $\mathbb {Z} \oplus \mathbb {Z}$ of two copies of the infinite cyclic group $\mathbb {Z}$. Symbolically,

$G=\{(a,b)|a,b\in \mathbb {Z} \}$.

One basis for this group is {(1,0),(0,1)}. If we say $\ e_{1}=(1,0)$ and $\ e_{2}=(0,1)$, then we can write the element (4,3) as

$\ (4,3)=4e_{1}+3e_{2}$. Where 'multiplication' is defined in following way: $\ 4e_{1}:=e_{1}+e_{1}+e_{1}+e_{1}$.

In this basis, there is no other way to write (4,3), but if we choose our basis to be {(1,0),(1,1)}, where $\ f_{1}=(1,0)$ and $\ f_{2}=(1,1)$, then we can write (4,3) as

$\ (4,3)=f_{1}+3f_{2}$.

Unlike vector spaces, 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.) The trivial abelian group {0} is also considered to be free abelian, with basis the empty set.

Every lattice forms a finitely-generated free abelian group.

Terminology

Note that a free abelian group is not a free group except in two cases: a free abelian group having an empty basis (rank 0, giving the trivial group) or having just 1 element in the basis (rank 1, giving the infinite 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, while in free abelian groups they must be identical. Free groups are the free objects in the category of groups, that is, the "most general" or "least constrained" groups with a given number of generators, whereas free abelian groups are the free objects in the category of abelian groups; in the general category of groups, it is an added constraint to require that ab = ba, whereas this is a required property in the category of abelian groups.

Properties

1. 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, where each function may take integer values, and all but finitely many of its values are zero. This is the direct sum of copies of $\mathbb {Z}$, one copy for each element of B.
2. 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.
3. 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.
4. 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.
5. Free abelian groups are a special case of free modules, as abelian groups are nothing but modules over the ring $\mathbb {Z}$.

Importantly, every subgroup of a free abelian group is free abelian (see below). 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.

It can be surprisingly difficult to determine whether a concretely given group is free abelian. Consider for instance the Baer–Specker group $\mathbb {Z} ^{\mathbb {N} }$, the direct product (not to be confused with the direct sum, which differs from the direct product on an infinite number of summands) of countably many copies of $\mathbb {Z}$. Reinhold Baer proved in 1937 that this group is not free abelian; Specker proved in 1950 that every countable subgroup of $\mathbb {Z} ^{\mathbb {N} }$ is free abelian.

Rank

Every finitely generated free abelian group is isomorphic to $\mathbb {Z} ^{n}$ 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.

Formal sum

A formal sum of elements of a given set B is an element of the free abelian group with basis B. In other words, given a set B, let G be the unique (up to isomorphism) free abelian group with basis B. For elements $b_{1},b_{2},...\in B$ and $a_{1},a_{2},...\in \mathbb {Z}$ (where there must be an $n\in \mathbb {Z}$ such that $a_{i}=0$ for all $i\geq n$),

$\sum _{i=1}^{\infty }a_{i}b_{i}\in G$

Subgroup closure

Every subgroup of a free abelian group is itself a free abelian group. This is similar 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 a free abelian group.

Proof: 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 $\mathbb {Z}$ (being nontrivial) and clearly free. Assume that if a group is generated by a set of size $\leq k$, 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 $\operatorname {pr} (a_{1}x_{1}+\dots +a_{k+1}x_{k+1})=a_{1}x_{1}.$ 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 nontrivial. 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 $y=nx+k$, where $k\in \operatorname {Ker} (\operatorname {pr} )$ and $n\in \mathbb {Z}$. 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 $\mathbb {Z}$.

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

$S=\{(G_{J},w)\mid G_{J}{\rm {\;is\;a\;nontrivial\;free\;group\;and\;}}w{\rm {\;is\;a\;basis\;of\;}}G_{J}\}.$

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

$w:J'\to G_{J}$

such that $w[J']$ generates $G_{J}$.

Clearly $S$ is nonempty: Let us have an element $x$ in $G$. Then $x=a_{1}x_{i_{1}}+\cdots +a_{n}x_{i_{n}}$ and thus the free group generated by $J=\{x_{i_{1}},\dots ,x_{i_{n}}\}$ contains $x$ and the intersection $G\cap F_{J}$ 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 $(G_{J},w)\leq (G_{K},u)$ 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 $u\upharpoonright w=w$.

If $(G_{J_{r}},w_{r})_{r\in L}$ is a $\leq$-chain ($L$ is some linear order) of elements of $S$, then obviously

$(\bigcup _{r\in L}G_{J_{r}},\bigcup _{r\in L}w_{r})\in S$,

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 $(G_{J},w)\leq (G_{K},w)$, 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 $n\in \mathbb {Z} \setminus \{0\}$ and $y\in F_{J}$.

The set of $n\in \mathbb {Z}$ for which there exists $y\in F_{J}$ such that $nx_{k}+y\in G_{K}$ forms a subgroup of $\mathbb {Z}$. Let $n_{0}$ be a generator of this group and let $w_{k}=n_{0}x_{k}+y\in G_{K}$ with $y\in F_{J}$. Now if $z\in G_{K}$, then for some $m\in Z$, $z=z-mw_{k}+mw_{k}$, where $z-mw_{k}\in G_{J}$.

On the other hand clearly $w_{k}\mathbb {Z} \cap F_{J}=\{0\}$, so $w'=w\cup \{w_{k}\}$ is a basis of $G_{K}$, so $(G_{K},w')\geq (G_{J},w)$ contradicting the maximality again. $\square$

References

• Phillip A. Griffith (1970). Infinite Abelian group theory. Chicago Lectures in Mathematics. University of Chicago Press. ISBN 0-226-30870-7.