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.])

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.