The strategy is to verify the three subgroup conditions — identity, closure, and inverses — directly, using the fact that each $H_i$ satisfies these conditions. Each verification reduces to checking membership in *every* $H_i$ separately, then combining the conclusions across the family. Associativity is inherited from $G$ automatically. **Step 1: Identity.** Since each $H_i \leq G$, every $H_i$ contains the identity $e_G$. Therefore $e_G \in \bigcap_{i \in I} H_i$. **Step 2: Closure.** Let $x, y \in \bigcap_{i \in I} H_i$. Then $x, y \in H_i$ for every $i \in I$. Since each $H_i$ is a subgroup, $xy \in H_i$ for every $i$. Therefore $xy \in \bigcap_{i \in I} H_i$. **Step 3: Inverses.** Let $x \in \bigcap_{i \in I} H_i$. Then $x \in H_i$ for every $i \in I$. Since each $H_i$ is a subgroup, $x^{-1} \in H_i$ for every $i$. Therefore $x^{-1} \in \bigcap_{i \in I} H_i$. Associativity is inherited from $G$. So $\bigcap_{i \in I} H_i \leq G$. The binary case $A \cap B \leq G$ recovers as $I = \{1,2\}$ with $H_1 = A$ and $H_2 = B$.