The proof treats each part separately. Part (i) verifies the homomorphism property for the composition using the homomorphism property of each factor. Part (ii) uses the bijective inverse and the homomorphism property of $\vartheta$ to derive the homomorphism property of $\vartheta^{-1}$. **Step 1: Composition of homomorphisms is a homomorphism.** Let $\varphi : A \to B$ and $\vartheta : B \to C$ be homomorphisms. For any $a, b \in A$: \begin{align*} (\vartheta \circ \varphi)(ab) = \vartheta(\varphi(ab)) = \vartheta(\varphi(a)\varphi(b)) = \vartheta(\varphi(a)) \cdot \vartheta(\varphi(b)) = (\vartheta \circ \varphi)(a) \cdot (\vartheta \circ \varphi)(b). \end{align*} So $\vartheta \circ \varphi$ is a homomorphism. If both $\varphi$ and $\vartheta$ are bijective (i.e., isomorphisms), then $\vartheta \circ \varphi$ is also bijective: it is injective because if $(\vartheta \circ \varphi)(a) = (\vartheta \circ \varphi)(b)$ then injectivity of $\vartheta$ gives $\varphi(a) = \varphi(b)$ and injectivity of $\varphi$ gives $a = b$; it is surjective because for any $c \in C$, surjectivity of $\vartheta$ gives $b \in B$ with $\vartheta(b) = c$, and surjectivity of $\varphi$ gives $a \in A$ with $\varphi(a) = b$, so $(\vartheta \circ \varphi)(a) = c$. Hence $\vartheta \circ \varphi$ is an isomorphism. Transitivity of $\cong$ follows: if $G_1 \cong G_2$ via $\varphi$ and $G_2 \cong G_3$ via $\vartheta$, then $G_1 \cong G_3$ via $\vartheta \circ \varphi$. **Step 2: Inverse of an isomorphism is an isomorphism.** Let $\vartheta : G_1 \to G_2$ be an isomorphism. Since $\vartheta$ is bijective, the inverse [function](/page/Function) $\vartheta^{-1} : G_2 \to G_1$ exists and is bijective. We must show $\vartheta^{-1}$ is a homomorphism. Let $a, b \in G_2$ and set $c = \vartheta^{-1}(a)$, $d = \vartheta^{-1}(b)$, so $\vartheta(c) = a$ and $\vartheta(d) = b$. Then: \begin{align*} \vartheta(cd) &= \vartheta(c)\vartheta(d) = ab. \end{align*} Applying $\vartheta^{-1}$ to both sides: \begin{align*} cd = \vartheta^{-1}(ab), \end{align*} i.e., $\vartheta^{-1}(ab) = \vartheta^{-1}(a) \cdot \vartheta^{-1}(b)$. So $\vartheta^{-1}$ is a homomorphism, and being bijective, it is an isomorphism.