Free Products

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Unlike in the case of abelian groups, the weak direct product defined in the previous post does not satisfy a universal property for general groups.

Example 1 Consider any nonabelian group \(G\), and let \(a,b\in G\) be such that \(ab\neq ba\). Define group homomorphisms \(f_1, f_2: (\mathbb{Z},+)\rightarrow (G,\cdot)\) by

\[f_1(1)=a, \qquad f_2(1)=b.\]

Since the index set \(I=\{1,2\}\) is finite, the weak direct product of two copies of \((\mathbb{Z},+)\) is simply \(\mathbb{Z}\times\mathbb{Z}\).

However, no map \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow G\) makes the following diagram commute:

Indeed, if such an \(f\) existed, we would have

\[\begin{aligned}ab&=f_1(1)f_2(1)=f(\iota_1(1))f(\iota_2(1))=f(\iota_1(1)+\iota_2(1))\\ &=f(\iota_2(1)+\iota_1(1))=f(\iota_2(1))f(\iota_1(1))=f_2(1)f_1(1)\\ &=ba\end{aligned}\]

contradicting the choice of \(a\) and \(b\).

Therefore, just as with direct sums for general groups, we need a new construction to obtain an object satisfying the universal property. To this end, we first define the free group.

Free group

Any group \(G\) can be regarded as a set equipped with a binary operation, an identity element, and inverses. Moreover, any group homomorphism can naturally be viewed as a function between sets. In other words, there is a forgetful functor \(U: \Grp \rightarrow\Set\). In this section we define the left adjoint \(F:\Set \rightarrow\Grp\) of \(U\). By the definition of a left adjoint functor, this is a functor satisfying the natural isomorphism

\[\Hom_\Set(X, U(G))\cong\Hom_\Grp(F(X), G).\]

(§Adjoint Functors, ⁋Definition 1) That is, for any set \(X\) and any group \(G\), the functor \(F\) yields a bijection that uniquely assigns to each \(f\in\Hom_\Set(X, U(G))\) an element of \(\Hom_\Grp(F(X),G)\). Rewriting this, we obtain the following.

Definition 2 For a nonempty set \(X\), the free group \(F(X)\) defined by \(X\) is determined as the solution \((F(X), \eta_X\in\Hom_\Set(X,UF(X)))\) of the following universal mapping problem.

For any group \(G\), if a function \(f:X\rightarrow U(G)\) is given, then there exists a unique group homomorphism \(\hat{f}:F(X)\rightarrow G\) satisfying \(U(\hat{f})\circ\eta_X=f\).

Here \(\eta_X\) is nothing but the unit of the adjunction \(F\dashv U\). Of course, to make use of this we must actually construct \(F(X)\).

We sketch the rough idea. First, consider a set \(X^{-1}\) disjoint from \(X\) and having the same cardinality. There is no reason for \(X^{-1}\) to be any special set, but we choose a bijection \(X\rightarrow X^{-1}\) and denote the image of \(x\in X\) in \(X^{-1}\) by \(x^{-1}\). Also, choose a singleton set disjoint from \(X\cup X^{-1}\) and let its single element be \(e\).

Then the elements of the group \(F\) are the collection of reduced words defined by the set \(X\cup X^{-1}\cup \{e\}\). Here a word is simply a list of elements from the set \(X\cup X^{-1}\cup \{e\}\); if the same element appears twice in a row, as in \(xx\), or if \(x^{-1}\) appears immediately after \(x\), as in \(xx^{-1}\), or if \(e\) lies between two terms, as in \(xey\), then these can be reduced to \(x^2\), \(e\), and \(xy\) respectively. However, if for example \(y\neq x^{-1}\), there is no way to reduce \(xyx\). A word that has been reduced as much as possible is called a reduced word.

Every word can be reduced to a reduced word.¹ Let us define the operation and identity element on these. The identity element is of course the reduced word \(e\). The operation is defined by simply concatenating two words and then reducing the result to a reduced word. For example, the product of the words \(x_1x_2\) and \(x_3x_4\) is given by \(x_1x_2x_3x_4\). Then \(e\) can also be viewed as the empty word under this operation. This operation is clearly associative. The inverse is obtained by taking the inverse of each term of the given element and then listing them in reverse order. For example, the inverse of the word

\[x_1x_2^{-1}x_3^2\]

is

\[x_3^{-2}x_2x_1^{-1}\]

and one can verify that the product of these two is indeed \(e\).

Now we have constructed the group \(F\), and if we identify the length-1 elements consisting of elements of \(X\) with the elements of \(X\) themselves, we also obtain \(\iota:X\rightarrow F\). Then one can easily show that these satisfy the universal property of Definition 2. To do so, we define \(\bar{f}\) as the function that replaces every element \(x\in X\) appearing in \(F\) with \(g(x)\), and then verify that this is a group homomorphism.

Corollary 3 Any group \(G\) is a homomorphic image of a free group.

Proof

Consider the set \(X\) of generators of \(G\), and then consider the free group \(F\) on \(X\). The function \(X\hookrightarrow G\) defines a group homomorphism from \(F\) to \(G\), and since the image of this homomorphism contains all the generators of \(G\), it is surjective.

Free product

Applying the same idea, we can similarly define the free product, which turns out to be the coproduct we have been seeking. Again, we only briefly sketch the construction.

Let a family of groups \((G_i)\) be given. For convenience, assume they are pairwise disjoint, and let \(X=\coprod G_i\). That is, for any element \(x\in X\), we can uniquely find an \(i\) such that \(x\in G_i\). Since the \(G_i\) already contain inverses, it suffices to take \(X\cup\{e\}\) as the set of generators.

The free product \(\prod^\ast G_i\) of \((G_i)\) is the set of reduced words formed from this set \(X\cup\{e\}\). The overall flow is the same as when defining a free group, but this time the elements of each \(G_i\) can be multiplied among themselves, so we must be somewhat more careful in defining reduced words. By a reduced word used in defining the free product, we mean a word

\[x_1x_2\cdots x_n\]

formed from elements of the set \(X\cup\{e\}\) satisfying the following three conditions:

1. If \(n>1\), none of the \(x_k\) equals \(e\).
2. If \(x_k\in X\), then \(x_k\) is not the identity element in the group \(G_i\) containing this element.
3. Any two adjacent elements \(x_i, x_{i+1}\) belong to different groups.

Given an arbitrary word, the way to reduce it is simple. Check whether adjacent elements belong to the same group; if they do, combine them into a single element by the operation in that group. If an identity element in some group appears during this process (or from the outset), simply erase that element.

Then the operation on \(\prod^\ast G_i\) is the same concatenation operation as when defining a free group, and it is not difficult to verify that this set carries a group structure. Moreover, a situation like Example 1 no longer occurs, because even if two groups \(G_1, G_2\) are abelian, their free product \(G_1\ast G_2\) is no longer an abelian group.

Example 4 Consider the same situation as Example 1. For notational convenience, let \(G_1=\langle a\rangle\cong\mathbb{Z}\) and \(G_2=\langle b\rangle\cong\mathbb{Z}\). Then the elements of \(G_1\ast G_2\) are the collection of elements of the form

\[ab, a^2b, a^{-1}ba^3, bab^2, \cdots\]

For example, multiplying the two elements \(a^2b\) and \(bab^2\) yields

\[(a^2b)(bab^2)=a^2bbab^2=a^2b^2ab^2\]

Here, \(\langle a\rangle\) and \(\langle b\rangle\) are cyclic subgroups of \(G_1\ast G_2\), so by defining homomorphisms from \(G_1\) and \(G_2\) into \(G_1\ast G_2\) via \(a\mapsto a\) and \(b\mapsto b\), we obtain the natural inclusion maps \(\iota_1\) and \(\iota_2\).

Of course, the same issue as in Example 1 does not arise. We have \(\iota_1(a)\iota_2(b)=ab\) and \(\iota_2(b)\iota_1(a)=ba\), and these two are distinct elements of \(\prod^\ast G_i\).

Proposition 5 The free product \(\prod^\ast G_i\) is the coproduct in \(\Grp\).

Proof

Let an arbitrary group \(H\) and group homomorphisms \(f_i:G_i\rightarrow H\) be given. Then by the universal property of \(X=\coprod U(G_i)\), there exists a unique function \(f:X\rightarrow U(H)\) satisfying \(U(f_i)=f\circ \iota_i\) for the inclusion maps \(\iota_i:U(G_i)\rightarrow X\). Now from the universal property in Definition 2, we obtain a group homomorphism \(\hat{f}:F(X)\rightarrow H\), and using the fact that the \(f_i\) are group homomorphisms, we see that \(f\) factors through the above reduction process and thus defines a map \(\prod^\ast G_i\rightarrow H\).

On the other hand, for an arbitrary group \(G\), a group homomorphism \(\mathbb{Z}\rightarrow G\) is uniquely determined by the image of \(1\in \mathbb{Z}\) in \(G\). That is, we have the isomorphism

\[\Hom_\Grp(\mathbb{Z},G)\cong U(G)\]

and by an argument similar to [Category Theory] §Representable Functors, we know that this isomorphism is a representation of \(U\); moreover, thinking of it as

\[\Hom_\Grp(\mathbb{Z},G)\cong \Hom_\Set(\ast, U(G))\]

we can interpret this as saying \(\mathbb{Z}=F(\ast)\). Therefore, for an arbitrary set \(X\), using [Category Theory] §Adjoints, ⁋Theorem 9, we can express the free group \(F(X)\) as a free product of copies of \(\mathbb{Z}\):

\(F(X)=F\left(\coprod_{x\in X} \{x\}\right)\cong \coprod_{x\in X} F(\ast)={\prod_{x\in X}}^\ast \mathbb{Z}\).

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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1. It is not strictly necessary to introduce reduced words in order to define the operation on words, but it is convenient to do so for the sake of uniqueness of representation. ↩