# 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.