The strategy is to show mutual divisibility between $o((h,k))$ and $\operatorname{lcm}(o(h), o(k))$ using the [Order Division Lemma](/theorems/771). **Step 1: $o((h,k)) \mid m$ where $m = \operatorname{lcm}(o(h), o(k))$.** Since $o(h) \mid m$ and $o(k) \mid m$, the [Order Division Lemma](/theorems/771) gives $h^m = e_H$ and $k^m = e_K$. Then: \begin{align*} (h, k)^m = (h^m, k^m) = (e_H, e_K). \end{align*} By the [Order Division Lemma](/theorems/771), $o((h,k)) \mid m$. **Step 2: $m \mid o((h,k))$.** Let $n = o((h,k))$. Then $(h,k)^n = (h^n, k^n) = (e_H, e_K)$, so $h^n = e_H$ and $k^n = e_K$. By the [Order Division Lemma](/theorems/771), $o(h) \mid n$ and $o(k) \mid n$, hence $m = \operatorname{lcm}(o(h), o(k)) \mid n$. **Step 3: Conclude.** Since $o((h,k)) \mid m$ and $m \mid o((h,k))$, we have $o((h,k)) = m$.