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.)


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.


  • Group

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