Free group

Intuitively, the free group \(F(X)\) on the set \(X\) is the group whose elements are the freely reduced words over \(X\), and whose group operation is “concatenation followed by free reduction”.

The free group can be constructed rigorously in several equivalent ways, some of which are easy to construct but hard to understand, and some of which are intuitive but rather hard to define properly. Our formal construction (detailed on the Formal Definition lens) will be one of the more opaque definitions; there, we eventually show that the formal construction yields a group which is isomorphic to the intuitive version, and this will prove that the intuitive version does in fact define a group with the right properties for the free group.

Intuitive definition

Given a set \(X\), the free group \(F(X)\) (or \(FX\)) on \(X\) has:


The free group on the set \(X = \{ a, b \}\) contains the following elements (among others):

  • The word \((a,b,a,a,a,b^{-1})\), which is written in shorthand as \(abaaab^{-1}\) or (most commonly) as \(aba^3b^{-1}\)

  • The empty word \(()\), which is written as \(\varepsilon\)

  • The word \((b,b,b)\), which is written as \(b^3\)

  • The word \((a^{-1}, b^{-1}, b^{-1})\), or \(a^{-1} b^{-2}\)

(From now on, we will only use the common shorthand to denote words, except in cases where this interferes with something else we’re doing.)

Some things which the same free group does not contain are:

  • \(aa^{-1}\), since this is not freely reduced

  • \(c\), for the fatuous reason that \(c\) is not a word over the alphabet \(\{a,b\}\)

  • \(abb^{-1}a\), since this is not freely reduced.

Some examples of this group’s group operation (which we will write as \(\cdot\)) in action are:

  • \(aba \cdot bab = ababab\)

  • \(aba^2 \cdot a^3b = aba^5b\)

  • \(aba^{-1} \cdot a = ab\), obtained as the free reduction of \(aba^{-1}a\)

  • \(ab \cdot b^{-1} a^{-1} = \varepsilon\), obtained as \(abb^{-1}a^{-1} = aa^{-1}\) (reduction of \(b\)) and then \(a a^{-1} = \varepsilon\).


  • The free group on the empty set is just the trivial group, the group with just one element. Indeed, we must have an identity (the empty word), but there are no extra generators to throw in, so there are no longer words.

  • The free group on a set with one element is isomorphic to the group of integers under addition. Indeed, say our set is \(\{ a \}\); then every member of the free group is a word of the form \(a^n\) or \(a^{-n}\) or \(a^0\). Then the isomorphism is that \(a^i\) is sent to \(i \in \mathbb{Z}\).

  • The free group on two elements is countable. Indeed, its elements are all of the form \(a^{i_1} b^{j_1} a^{i_2} b^{j_2} \dots a^{i_n} b^{j_n}\), and so the free group injects into the set of natural numbers by the (rather convoluted) \($a^{i_1} b^{j_1} a^{i_2} b^{j_2} \dots a^{i_n} b^{j_n} \mapsto 2^{\mathrm{sgn}(i_1)+2} 3^{|i_1|} 5^{\mathrm{sgn}(j_1)+2} 7^{|j_1|} \dots\)$ where we produce the natural number noteUniquely, by the fundamental theorem of arithmetic. whose prime factors tell us exactly how to construct the original word. noteThis trick is extremely useful in determining whether a set is countable, and you should go over it until you are sure you understand it. (The \(\mathrm{sgn}\) function outputs \(-1\) if its input is negative; \(1\) if the input is positive; and \(0\) if its input is \(0\).) As a related exercise, you should prove that the free group on a countable set is countable.

Why are the free groups important?

It is a fact that every group is a quotient of a free group. Therefore the free groups can be considered as a kind of collection of “base groups” from which all other groups can be made as quotients. This idea is made more concrete by the idea of the group presentation, which is a notation that specifies a group as a quotient of a free group. noteAlthough it’s not usually presented in this way at first, because the notation has a fairly intuitive meaning on its own.

Free groups “have no more relations than they are forced to have”

The crux of the idea of the group presentation is that the free group is the group we get when we take all the elements of the set \(X\) as elements which “generate” our putative group, and then throw in every possible combination of those “generators” so as to complete it into a bona fide group. We explain why the free group is (informally) a “pure” way of doing this, by walking through an example.

If we want a group which contains the elements of \(X = \{ a, b \}\), then what we could do is make it into the cyclic group on the two elements, \(C_2\), by insisting that \(a\) be the identity of the group and that \(b \cdot b = a\). However, this is a very ad-hoc, “non-pure” way of making a group out of \(X\). %%note:For those who are looking out for that sort of thing, to do this in generality will require heavy use of the axiom of choice.%% It adds the “relation” \(b^2 = a\) which wasn’t there before.

Instead, we might make the free group \(FX\) by taking the two elements \(a, b\), and throwing in everything we are forced to make if no non-trivial combination is the identity. For example:

  • we have no reason to suspect that either \(a\) or \(b\) is already the identity, so we must throw in an identity which we will label \(\varepsilon\);

  • we have no reason to suspect that \(a \cdot b\) should be equal to \(a\) or to \(b\) or to \(\varepsilon\), so we must throw in an element equal to \(a \cdot b\) which we will label \(ab\);

  • we do have a reason to suspect that \(a^{-1} \cdot a\) is already in the group (because it “ought to be” the identity \(\varepsilon\)), so we don’t throw in the element \(a^{-1} a\);

  • and so on through all the words, such as \(a^{-1}ba^2b^{-2}\) and so forth.

This way adds an awfully large number of elements, but it doesn’t require us to impose any arbitrary relations. <div><div>

Universal property of the free group

The free group has a universal property, letting us view groups from the viewpoint of category theory. Together with the idea of the quotient (which can be formulated in category theory as the coequaliser) and the subsequent idea of the group presentation, this lets us construct any group in a category-theoretic way.

Indeed, every group \(G\) has a presentation \(\langle X \mid R \rangle\) say, which expresses \(G\) as a quotient (i.e. a certain coequaliser) of the free group \(F(X)\). We can construct \(F(X)\) through the universal property, so we no longer need to say anything at all about the elements or group operation of \(G\) to define it.


  • It is not obvious, but \(FX\) is isomorphic to \(FY\) if and only if \(X\) and \(Y\) have the same cardinality as sets. (Proof.) This completely classifies the free groups and tells us just when two free groups are “the same”, which is nice to have.

  • The only two abelian groups which are free are the trivial group (of one element; i.e. the free group on no generators), and \(\mathbb{Z}\) the free group on one generator. No others are abelian because if there are two or more generators, pick \(a, b\) to be two of them; then \(ab \not = ba\) by free-ness. noteTo be completely precise about this, using the formal definition, \(\rho_a \rho_b \not = \rho_b \rho_a\) because they have different effects when applied to the empty word \(\varepsilon\). One produces \(ab\); the other produces \(ba\).

  • The free groups are all torsion-free note”Torsion-free” means “free of torsion”, where the word “free” is nothing to do with “free group”.: that is, no element has finite order unless it is the identity \(\varepsilon\). (Proof.) This property helps us to recognise when a group is not free: it is enough to identify an element of finite order. However, the test doesn’t go the other way, because there are torsion-free groups which are not free.

Suppose the order of an element \(x \in \mathbb{Q}\) were finite and equal to \(n \not = 0\), say. Then \(x+x+\dots+x\), \(n\) times, would yield \(0\) (the identity of \((\mathbb{Q}, +)\)). But that would mean \(n \times x = 0\), and so \(n=0\) or \(x = 0\); since \(n \not = 0\) already, we must have \(x = 0\), so \(x\) is the identity after all.

The group is not free: suppose it were free. It is abelian, so it must be isomorphic to either the trivial group (clearly not: \(\mathbb{Q}\) is infinite but the trivial group isn’t) or \(\mathbb{Z}\). It’s not isomorphic to \(\mathbb{Z}\), though, because \(\mathbb{Z}\) is cyclic: there is a “generating” element \(1\) such that every element of \(\mathbb{Z}\) can be made by adding together (possibly negatively-many) copies of \(1\). \(\mathbb{Q}\) doesn’t have this property, because if \(x\) were our attempt at such a generating element, then \(\frac{x}{2}\) could not be made, so \(x\) couldn’t actually be generating after all. <div><div>



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