The free-group monad has groups as its Eilenberg–Moore algebras

By [L1]–[L2], the monad sends $X$ to the underlying set of the free group $F(X)$, its unit inserts generators, and its multiplication evaluates a reduced word whose letters are themselves reduced words. Every group $H$ gives an algebra $F(UH)\to UH$ by word evaluation.

Conversely, from an algebra $a:UF(X)\to X$, define the product, identity, and inverse by evaluating the free-group words $xy$, $1$, and $x^{-1}$. The algebra unit law fixes generators, while the multiplication law identifies evaluation after substitution with direct evaluation; applying it to the standard group-word identities gives all axioms in [L3].

An algebra homomorphism commutes with word evaluation and therefore preserves product, identity, and inverse. Conversely a group homomorphism preserves every group word and hence is an algebra homomorphism. The two constructions are inverse over $\mathbf{S}\mathbf{e}\mathbf{t}$.