[proofplan] The subgroup hypothesis says that $H$ is already a group under the restriction of the operation of $G$. Thus only commutativity has to be checked. We take arbitrary elements of $H$, view them as elements of $G$, use that $G$ is abelian, and then interpret the resulting equality back inside the inherited operation on $H$. [/proofplan] [step:Use the subgroup structure inherited from $G$] Let $\cdot_H: H \times H \to H$ denote the operation on $H$ inherited from $G$, so for all $a,b \in H$ one has $a \cdot_H b = a \cdot b$ as elements of $G$. Since $H \le G$, the set $H$ equipped with $\cdot_H$ is a group. [guided] The notation $H \le G$ means that $H$ is a subgroup of $G$. In particular, $H$ is not merely a subset: it is a group under the operation inherited from $G$. Let us name that inherited operation explicitly: \begin{align*} \cdot_H: H \times H \to H. \end{align*} By definition of inherited operation, for every $a,b \in H$, the product $a \cdot_H b$ is the same element as the product $a \cdot b$ computed in $G$. The subgroup hypothesis also gives the group axioms for $(H,\cdot_H)$, so to prove that $H$ is abelian we only need to prove that $\cdot_H$ is commutative. [/guided] [/step] [step:Restrict commutativity from $G$ to $H$] Let $h_1,h_2 \in H$ be arbitrary. Since $H \subset G$, we have $h_1,h_2 \in G$. Because $(G,\cdot)$ is abelian, \begin{align*} h_1 \cdot h_2 = h_2 \cdot h_1. \end{align*} Using the definition of $\cdot_H$ on both sides gives \begin{align*} h_1 \cdot_H h_2 = h_2 \cdot_H h_1. \end{align*} Since $h_1$ and $h_2$ were arbitrary elements of $H$, the operation $\cdot_H$ is commutative. Therefore $(H,\cdot_H)$ is an [abelian group](/page/Abelian%20Group). [/step]