The strategy is to use the [Order Division Lemma](/theorems/771) in both directions — applied through the isomorphism and its inverse — to show mutual divisibility, which forces equality. **Step 1: $o(b) \mid o(a)$.** Let $n = o(a)$. Then $a^n = e_G$. Applying $\vartheta$: \begin{align*} e_H = \vartheta(a^n) = \vartheta(a)^n = b^n. \end{align*} By the [Order Division Lemma](/theorems/771), $o(b) \mid n = o(a)$. **Step 2: $o(a) \mid o(b)$.** Let $m = o(b)$. Then $b^m = e_H$. Since $\vartheta$ is an isomorphism, $\vartheta^{-1}$ is also an isomorphism by the [Functional Properties of Isomorphisms](/theorems/770). Applying $\vartheta^{-1}$: \begin{align*} e_G = \vartheta^{-1}(b^m) = \vartheta^{-1}(b)^m = a^m. \end{align*} By the [Order Division Lemma](/theorems/771), $o(a) \mid m = o(b)$. **Step 3: Conclude.** Since $o(a) \mid o(b)$ and $o(b) \mid o(a)$ with both positive, $o(a) = o(b)$.