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  text: 'Given a [3gd group] $G$, there is a [-free_group] $F(X)$ on some set $X$, such that $G$ is [49x isomorphic] to some [4tq quotient] of $F(X)$.\n\nThis 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.])\n\n# Proof\nLet $F(G)$ be the free group on the elements of $G$, in a slight abuse of notation where we use $G$ interchangeably with its [-3gz].\nDefine the [47t homomorphism] $\\theta: F(G) \\to G$ by "multiplying out a word": taking the word $(a_1, a_2, \\dots, a_n)$ to the product $a_1 a_2 \\dots a_n$.\n\nThis 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 $$\\theta(w_1 w_2) = \\theta(a_1, \\dots, a_m, b_1, \\dots, b_m) = a_1 \\dots a_m b_1 \\dots b_m = \\theta(w_1) \\theta(w_2)$$\n\nThis immediately expresses $G$ as a quotient of $F(G)$, since [4h7 kernels of homomorphisms are normal subgroups].',
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