Given a group \(G\), there is a free group \(F(X)\) on some set \(X\), such that \(G\) is isomorphic to some quotient of \(F(X)\).

This is an instance of a much more general phenomenon: for a general monad \(T: \mathcal{C} \to \mathcal{C}\) where \(\mathcal{C}\) is a category, if \((A, \alpha)\) is an algebra over \(T\), then \(\alpha: TA \to A\) is a coequaliser. (Proof.)

Proof

Let \(F(G)\) be the free group on the elements of \(G\), in a slight abuse of notation where we use \(G\) interchangeably with its underlying set. Define the homomorphism \(\theta: F(G) \to G\) by “multiplying out a word”: taking the word \((a_1, a_2, \dots, a_n)\) to the product \(a_1 a_2 \dots a_n\).

This is indeed a group homomorphism, because the group operation in \(F(G)\) is concatenation and the group operation in \(G\) is multiplication: clearly if \(w_1 = (a_1, \dots, a_m)\), \(w_2 = (b_1, \dots, b_n)\) are words, then \($\theta(w_1 w_2) = \theta(a_1, \dots, a_m, b_1, \dots, b_m) = a_1 \dots a_m b_1 \dots b_m = \theta(w_1) \theta(w_2)\)$

This immediately expresses \(G\) as a quotient of \(F(G)\), since kernels of homomorphisms are normal subgroups.