# What is the definition of a free abelian group generated by the set $X$?

What is the definition of a free abelian group generated by the set $X$?

If $X=\{A,B,C,D\}$ what are the elements of the free abelian group on this set? What is the identity of this group?

• Which definition have you seen? Jan 31, 2018 at 12:40
• I don't have a specific definition. I've come across the expression "free abelian group generated by a set $X$" and I'm trying to find an explicit definition for it. Jan 31, 2018 at 13:01
• Then you should probably ask about that, rather than about understanding the definition. But the definition can be found on Wikipedia for example. Jan 31, 2018 at 13:02
• You could take a look at free abelian group at wiki. Jan 31, 2018 at 13:04
• I will edit my question to better reflect that question. Jan 31, 2018 at 13:05

The free abelian group $G(X)$ generated by $X = \{A,B,C,D\}$ is the smallest abelian group containing $X$ and has no other restricting properties. You should think of $X$ as a basis for the group $G(X)$, similarly to a basis for a vector space.

Concretely, what are the elements of $G(X)$? Starting from $X$, first adjoin symbols that will represent the identity in $G(X)$ and inverses for the generating elements: $0, −A, -B, -C, -D$. Now form the set of all abstract finite sums, e.g. $A + B + D + (-B)$. Finally, introduce relations (i.e. define the set of equivalence classes of sums) saying that these elements and sums behave like a commutative group; for example, $A + B + D + (-B) = A + D$.

Since the group is abelian, we can collect terms; for instance, $A + B + A = A + A + B = 2A + B$. So, each element of the free abelian group $G(X)$ can be expressed uniquely as a sum $$n_AA + n_BB + n_CC + n_DD$$ where each $n_x \in \mathbb{Z}$; in other words, $G(X) = \big\{\sum_{x \in X} n_X X \mid n_x \in \mathbb{Z} \big\}$. Therefore, you can easily see that $G(X) \simeq \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$, where the group operation is the pointwise addition and the neutral element is $(0,0,0,0)$.

In the explanation above, note that actually the nature of the elements of $X$ plays no role, the only relevant information is the number of elements of $X$. Thus, more generally, you can see that given a set $X$ with exactly $n \in \mathbb{N}$ elements, $G(X) \simeq \underbrace{\mathbb{Z} \times \dots \times \mathbb{Z}}_{n \text{ times } \mathbb{Z}}$.

• So $G(X) = \{\sum_{x \in X}n_x x : n_x \in \Bbb Z\}$ with pointwise addition as the operation? What's the identity? Jan 31, 2018 at 13:21
• Yes! The identity is then $0A+0B+0C+0D$ i.e. $(0,0,0,0)$ if you consider that $G(X)≃ℤ×ℤ×ℤ×ℤ$. Jan 31, 2018 at 13:24

It is the abelian group $$\mathbf Z^{(X)}$$ i.e. the set of functions $$f:X\rightarrow \mathbf Z$$ with finite support. If, as is usual, we represent such a function by its values at each point $$x\in X$$, it is $$\mathbf Z^{(X)}=\Bigl\{(n_x)_{x\in X}\mid n_x=0\enspace\text{except for a finite number of them}\Bigr\}.$$

If $$|X|=n$$, it'is the same as $$\mathbf Z^X=\displaystyle\prod_{x\in X}\mathbf Z$$ and it's isomorphic to $$\mathbf Z^n$$.

Note: as an abelian group, the ring of polynomials with integer coefficients is defined as $$\mathbf Z^{(\mathbf N)}$$, to which one adds an extra multiplicative structure. In contrast, $$\mathbf Z^{\mathbf N}$$ with the extra multiplicative structure given by the Cauchy product, is the ring of formal power series with integer coefficients.

One way you can think of this in a very elementary and canonical way is that the free abelian group on a finite set such as $\{A,B,C,D\}$ is the group of all functions $$f : \{A,B,C,D\} \to \mathbb{Z}$$ with the operations of pointwise addition and pointwise inverse: given any two such functions, for all $x \in \{A,B,C,D\}$ define $$(f_1+f_2)(x) = f_1(x) + f_2(x)$$ also $$(-f)(x) = -f(x)$$ The identity element, of course, is the constant function $f(x)=0$.

For infinite sets, you have to add one more clause to the definition: for the free abelian group on a set $X$, you do not take all functions $f : X \to \mathbb{Z}$, you take only those functions which satisfy the property that there is a finite subset $Y \subset X$ such that if $x \in X-Y$ then $f(x)=0$.

I preassume that here the elements $A,B,C,D$ are distinct.

An abelian group free on set $\{A,B,C,D\}$ is the group $\langle\mathbb Z^4,+\rangle$ where $+:\mathbb Z^4\times\mathbb Z^4\to\mathbb Z^4$ is prescribed by:$$+(\langle n_A,n_B,n_C,n_D\rangle,\langle m_A,m_B,m_C,m_D\rangle)=\langle n_A+m_A,n_B+m_B,n_C+m_C,n_D+m_D\rangle$$ Or more commonly:$$\langle n_A,n_B,n_C,n_D\rangle+\langle m_A,m_B,m_C,m_D\rangle=\langle n_A+m_A,n_B+m_B,n_C+m_C,n_D+m_D\rangle$$

If $\eta:\{A,B,C,D\}\to\mathbb Z^4$ is prescribed by:

• $A\mapsto\langle1,0,0,0\rangle$
• $B\mapsto\langle0,1,0,0\rangle$
• $C\mapsto\langle0,0,1,0\rangle$
• $D\mapsto\langle0,0,0,1\rangle$

Then for every function $f:\{A,B,C,D\}\to A$ where $A$ denotes an abelian group there is a unique groupmorphism $\phi:\langle\mathbb Z^4,+\rangle\to A$ such that $f=\phi\circ\eta$.

This universal property is characteristic for a "free abelian group on a set".

More explicitly: $$\phi(n_A,n_B,n_C,n_D)=n_Af(A)+n_Bf(B)+n_Cf(C)+n_Df(D)$$

Here e.g. $3a$ where $a\in A$ must be looked at as a notation for $a+a+a$.

• So a free abelian group on a set $X$ is defined as: $\langle \Bbb Z^k, + \rangle$ where $k=|X|$ is the cardinality of $X$? As in the content of $X$ is (isomorphically) irrelevant? Jan 31, 2018 at 13:28
• That works indeed if the cardinality of the set is finite. Also if abelian group $A$ serves as free abelian group over some set then abelian group $B$ will serve as free abelian group over that set if and only if $A$ and $B$ are isomorphic. Jan 31, 2018 at 13:36

Long story short: it is the set of all words in the four letters $A, B, C, D$ and their formal inverses $A^{-1}, B^{-1}, C^{-1}, D^{-1}$, where any two letters commute.

E.g. $ABBADC^{-1}$ is one such word, and by commutation it is equivalent to $AABBC^{-1}D$. It is not hard to see why this group is isomorphic to $\mathbb{Z}^4$ (via the powers in the notation $A^2B^2C^{-1}D^1$).

"Free" in this context is here to emphasise that there are no other relations than the commutation.

Other answers have already pointed out concrete models for free Abelian groups. I want to emphasize that they can be characterized by a universal property, which I found helpful to gain some perspective on the construction.

In order to state the universal property, note that every Abelian group $$G$$ has an underlying set $$\underline G$$. The passage from $$G$$ to $$\underline G$$ simply forgets the group structure. Similarly, a group homomorphism $$\alpha\colon G\to H$$ has an underlying map of sets $$\underline \alpha\colon \underline G\to\underline H$$.

Now let $$S$$ be a set. An Abelian group $$F(S)$$ together with a map $$i\colon S\to \underline{F(S)}$$ is called freely generated by $$S$$ if for every Abelian group $$G$$ and every map $$a\colon S\to \underline G$$ there is a unique group homomorphism $$\alpha\colon F(S)\to G$$ such that $$a=\underline \alpha\circ i$$. In other words, the map $$i$$ gives a bijection $$Hom(F(S),G) \cong map(S,\underline G),\quad \alpha\mapsto \underline \alpha\circ i.$$ This property characterizes the pair $$(F(S),i)$$ uniquely up to unique isomorphisms. It manifests the slogan that "homomorphisms out of $$F(S)$$ are determined by the values on $$S$$".

EDIT: By the way, if you erase the word "Abelian" everywhere, you get the universal property of free groups. In fact, I initially missed that word and wrote the answer for groups. Even more generally, you can speak of "free objects" in other categories.

• Typo: $\alpha$ should go from $G$ to $H$, not $G$ to $G$. Feb 21, 2021 at 17:46
• @RaviFernando Gone. Thanks. Feb 21, 2021 at 18:00