[proofplan] We compare the two homomorphisms by collecting all elements of $G$ on which they agree. This agreement set is a subgroup of $G$ because homomorphisms preserve identity, multiplication, and inverses. Since it contains the generating set $S$, the defining minimality of $\langle S\rangle$ forces it to contain all of $G$, which is exactly the statement that $\varphi=\psi$. [/proofplan] [step:Form the subgroup on which the two homomorphisms agree] Let $1_G$ denote the identity element of $G$, and let $1_H$ denote the identity element of $H$. Define the agreement set $A \subset G$ by \begin{align*} A := \{g \in G : \varphi(g)=\psi(g)\}. \end{align*} We prove that $A \le G$. First, since $\varphi,\psi: G \to H$ are [group homomorphisms](/page/Group%20Homomorphism), they preserve identity elements, so \begin{align*} \varphi(1_G)=1_H=\psi(1_G). \end{align*} Thus $1_G \in A$. Next, let $a,b \in A$. Then $\varphi(a)=\psi(a)$ and $\varphi(b)=\psi(b)$. Using the homomorphism property for $\varphi$ and $\psi$, and then substituting these two equalities, we get \begin{align*} \varphi(ab) &=\varphi(a)\varphi(b) \\ &=\psi(a)\psi(b) \\ &=\psi(ab). \end{align*} Thus $ab \in A$. Finally, let $a \in A$. Then $\varphi(a)=\psi(a)$. Since homomorphisms preserve inverses, \begin{align*} \varphi(a^{-1}) &=\varphi(a)^{-1} \\ &=\psi(a)^{-1} \\ &=\psi(a^{-1}). \end{align*} Thus $a^{-1} \in A$. Therefore $A$ is a [subgroup](/page/Subgroup) of $G$. [guided] The purpose of defining $A$ is to turn equality of two functions into a structural statement inside the domain group. Define \begin{align*} A := \{g \in G : \varphi(g)=\psi(g)\}. \end{align*} To prove $A \le G$, we verify the subgroup conditions: identity, closure under multiplication, and closure under inverses. Let $1_G$ be the identity element of $G$, and let $1_H$ be the identity element of $H$. Because $\varphi,\psi: G \to H$ are [group homomorphisms](/page/Group%20Homomorphism), each sends the identity of $G$ to the identity of $H$. Hence \begin{align*} \varphi(1_G)=1_H=\psi(1_G), \end{align*} so $1_G \in A$. Now take arbitrary elements $a,b \in A$. By the definition of $A$, this means \begin{align*} \varphi(a)=\psi(a) \quad\text{and}\quad \varphi(b)=\psi(b). \end{align*} We must show $ab \in A$, which means showing $\varphi(ab)=\psi(ab)$. The homomorphism property gives \begin{align*} \varphi(ab)=\varphi(a)\varphi(b) \quad\text{and}\quad \psi(ab)=\psi(a)\psi(b). \end{align*} Substituting $\varphi(a)=\psi(a)$ and $\varphi(b)=\psi(b)$ into the first product gives \begin{align*} \varphi(ab) &=\varphi(a)\varphi(b) \\ &=\psi(a)\psi(b) \\ &=\psi(ab). \end{align*} Thus $ab \in A$. Finally, take arbitrary $a \in A$. Again, by definition of $A$, we have $\varphi(a)=\psi(a)$. We must show $a^{-1} \in A$, meaning $\varphi(a^{-1})=\psi(a^{-1})$. Since group homomorphisms preserve inverses, \begin{align*} \varphi(a^{-1})=\varphi(a)^{-1} \quad\text{and}\quad \psi(a^{-1})=\psi(a)^{-1}. \end{align*} Using $\varphi(a)=\psi(a)$, their inverses in $H$ are equal, so \begin{align*} \varphi(a^{-1}) &=\varphi(a)^{-1} \\ &=\psi(a)^{-1} \\ &=\psi(a^{-1}). \end{align*} Therefore $a^{-1} \in A$. We have shown that $A$ contains $1_G$, is closed under products, and is closed under inverses, so $A$ is a [subgroup](/page/Subgroup) of $G$. [/guided] [/step] [step:Use the generating property to force agreement on all of $G$] By hypothesis, $\varphi(s)=\psi(s)$ for every $s \in S$. Therefore $S \subset A$. Since $A \le G$ and $A$ contains $S$, the defining minimality of the [generated subgroup](/page/Generated%20Subgroup) gives \begin{align*} \langle S\rangle \subset A. \end{align*} The hypothesis $G=\langle S\rangle$ then gives $G \subset A$. Since $A \subset G$ by definition, we have $A=G$. [/step] [step:Conclude that the two homomorphisms are equal] Because $A=G$, every $g \in G$ belongs to $A$. By the definition of $A$, this means \begin{align*} \varphi(g)=\psi(g) \end{align*} for every $g \in G$. Hence $\varphi=\psi$ as functions $G \to H$. [/step]