Every group is a quotient of a free group


by Patrick Stevens Jul 22 2016

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_category_theory monad] $~$T: \mathcal{C} \to \mathcal{C}$~$ where $~$\mathcal{C}$~$ is a category, if $~$(A, \alpha)$~$ is an [eilenberg_moore_category algebra] over $~$T$~$, then $~$\alpha: TA \to A$~$ is a [coequaliser_category_theory coequaliser]. ([algebras_are_coequalisers 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.