# Every group is a quotient of a free group

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.

Parents:

• Group

The algebraic structure that captures symmetry, relationships between transformations, and part of what multiplication and addition have in common.