Given a group , 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 $T : C \to C$ where $C$ is a category, if $(A, α)$ is an algebra over $T$ , then $α : T A \to A$ 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 $θ : F (G) \to G$ 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 $θ (w_{1} w_{2}) = θ (a_{1}, \dots, a_{m}, b_{1}, \dots, b_{m}) = a_{1} \dots a_{m} b_{1} \dots b_{m} = θ (w_{1}) θ (w_{2})$

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