> [!NOTE] Theorem > Let $G=\langle S \mid R\rangle$ be a [[Group Presentation|group presentation]] where $S=\{ s_{1},s_{2},\dots, s_{n} \}$ is a [[Finite Set|finite]] set of generators and $R$ is a set of relations. Let $H$ be a [[Groups|group]] and $h_{1},h_{2},\dots,h_{n}\in H$. > > There exists a [[Homomorphisms of groups|homomorphism]] $\phi:G\to H$ satisfying $\phi(s_{i})=h_{i}$ iff every relation $r\in R$ holds with the $s_{i}$ replaced by the $h_{i}.$ Moreover, in this case the homomorphism is unique. ###### Proof sketch First, note that G can be constructed as the quotient of the free group F(S) on the generator set S=\{s_{1},\dots,s_{n}\} by the normal subgroup N generated by the relators R.  By the universal property of the free group, any assignment of the generators s_{i} to elements h_{i}\in H extends uniquely to a homomorphism \tilde\phi: F(S)\to H.  This map factors through the quotient F(S)/N precisely when every relator r\in R is sent to the identity in H, i.e.\ when N\subseteq\ker\tilde\phi.  In that case there is a unique induced homomorphism \phi: G=F(S)/N\to H with \phi(s_{i})=h_{i}.  Conversely, if such a \phi exists then its composition with the projection F(S)\to G gives a lift \tilde\phi whose kernel contains N, so all relators must map to the identity.  Uniqueness of \phi follows because every element of G is represented by a word in the s_{i}, and its image is forced by the images of those generators.