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The free group

April 16, 2012

A free object (say a group, an algebra or a topological space) over a set S can be thought of as the generic object of that type containing a copy of S. In other words, the only relations between the elements of the free object are the relations needed for it to satisfy it’s defining properties. Even if you never heard about the notion of a free object explicitly, the idea should be familiar. For example, the free vector space over a set S is simply the vector space with basis S. After constructing the free group, speaking of “the” free object over S will be shown to be justified. Indeed, as a consequence of the fact that free objects satisfy a so-called universal property, which could thus be used as a definition, they are unique up to isomorphism.

As an illustrative example of a free object, we’ll construct the free group over a set S. The idea will be to consider the set of all words in a well-chosen alphabet together with the operation of concatenation of words, and then to impose the minimal relations between these words in order for this set to be a group. Take a set S^{-1} in bijection with S but disjoint from it. For convenience, for each s \in S, let’s note s^{-1} \in S^{-1} it’s corresponding element and vice versa (so that (s^{-1})^{-1} = s). Our last ingredient will be a singleton {1} with 1 \notin S \cup S^{-1} and 1^{-1} = 1. Note A(S) the union of these 3. In other words, A(S) is the disjoint union of S with itself and a singleton. Now denote W(S) the set of all finite words on the alphabet A(S), i.e.

W(S) = \{ a_1a_2\ldots a_n \mid a_i \in A(S) ; n \in \textbf{N} \}.

Define the concatenation of words * on the alphabet A by

(a_1\ldots a_r)*(b_1 \ldots b_s) = a_1\ldots a_r b_1 \ldots b_s

where a_i, b_i \in A. Clearly W(S) is closed under concatenation of words on A(S). It only remains to give it a group structure by introducing some relations between some of its elements. But the key idea in the notion of a free object, is to introduce only the minimal relations. In this case, we want an identity element and an inverse for every element. This will be achieved minimally with the following relation :

1w \sim w1 \sim w;

ww^{-1} \sim w^{-1}w \sim 1.

The resulting set of equivalence classes with the obvious operation induced by concatenation of words, noted (W(S)/\sim, *), is what is called a free group over S and is often denoted by F(S). It would now only remain to verify that F(S) is indeed a group, which I will not do.

Note that the set S can be viewed as being contained in F(S) by associating the 1-letter word s to every s \in S. This inclusion map will be denoted by \iota.

A very important property of free groups is that they satisfy the following universal property:

Theorem: (universal property) Let G be a group, S a set and \varphi : S \to G a map from S to G. Then there exists a unique group homomorphism \phi : F(S) \to G such that \phi(s) = \varphi (\iota (s)) for all s \in S. i.e. this diagram commutes :

Proof: Let \phi : F(S) \to G be defined on \iota(S) with \phi(s) = \varphi(s) for all s \in \iota(S). Then, since S generates F(S), there is a unique way of extending \phi to a homomorphism defined on all of F(S). Namely, we need to have

\phi(s_1^{\epsilon_1} \ldots s_n^{\epsilon_n}) = \phi(s_1)^{\epsilon_1} \ldots \phi(s_n)^{\epsilon_n}

where \epsilon_i \in \{-1,1\} and s is identified with s^1. Note that this is the same idea as in linear algebra where a linear map is entirely determined by it’s action on a set of generators for the vector space.

QED

So, why is this property so important? Well, it guarantees that the free group over S is (up to isomorphism) the most general group containing S. In a sense, it guarantees us that if what we want is a master group containing all the information about the set S (so that we lose nothing in passing from S to our generic group), this is exactly what we got. This property could even be used as the definition of a free group :

Corollary: All groups satisfying the universal property of free groups (satisfying the theorem) are isomorphic.

Proof: Suppose F(S) and F'(S) are two groups satisfying the universal property, in other words, they are two free groups generated by S. Denote by \iota_1 (resp \iota_2) the natural injection from S to F(S) (resp F'(S)) and identify S with its image by \iota_1, \iota_2 in F(S), F'(S). Then by the universal property, there exists unique group homomorphisms \phi : F(S) \to F'(S) and \phi' : F'(S) \to F(S) such that \phi|S = \phi'|S = Id. Since the composition \phi' \circ \phi is in End(F(S)) and is the identity on S, by uniqueness it must be the identity map. Similarly, \phi \circ \phi' =  Id thus \phi is an isomorphism.

QED

Finally, for any group G, consider the identity homomorphism of G as a set map from G to G. Then there exists a unique \phi : F(G) \to G such that \phi(G) = G so \phi is surjective. By the first isomorphism theorem, this shows that G is isomorphic to the quotient of F(G) by \ker \phi. This is the idea behind the presentation of groups by generators and relations : every finitely generated group is constructible as a quotient of a free group by some relations.

Reference: D. S. Dummit, R. M. Foote, Abstract Algebra, 3rd ed.

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From → Algebra, Tensor, symmetric, exterior algebra

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