Given a group $G$, there is a free_group $F\left(X\right)$ on some set $X$, such that $G$ is isomorphic to some quotient of $F\left(X\right)$.

This is an instance of a much more general phenomenon: for a general monad $T:C\to C$ where $C$ is a category, if $(A,\alpha )$ is an algebra over $T$, then $\alpha :TA\to A$ is a coequaliser. (Proof.)

Proof

Let $F\left(G\right)$ 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 $\theta :F\left(G\right)\to G$ 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 $F\left(G\right)$ 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 $$\theta \left(w_1{w}_2\right)=\theta (a_1,\dots ,a_m,b_1,\dots ,b_m)=a_1\dots a_m{b}_1\dots b_m=\theta \left(w_1\right)\theta \left(w_2\right)$$

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