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Problem 72 (An External Direct Product Of Subgroups Is A Subgroup Of An External Direct Product)

Show that if $A$ is a subgroup of $G$ and $B$ is a subgroup of $H$, then $A\times B$ is subgroup of $G\times H$.

Solution

Let $G$ and $H$ be groups. Let $A\leq G$ and let $B\leq H$. We will use the definition of a subgroup to prove $A\times B$ is a subgroup of $G\times H$. By definition, we know $A$ and $B$ are nonempty, which means $A\times B$ is nonempty. We must show $A\times B \subseteq G\times H$. Recall $A\times B = \{(a,b) | a\in A, b\in B\}$. We will pick $(a,b)\in A\times B$. Since $a\in A$ and $A\leq G$, we know $a\in G$. Similarly, since $b\in B$ and $B\leq H$, we know $b\in H$. Thus $(a,b)\in G\times H$.

Since $A$ and $B$ are groups, we know $A\times B$ is a group (See exercise prior to this problem). Since $A\times B$ is a group we know it is closed under the group operation.

By the definition of a subgroup, we know $A\times B \leq G\times H$. $\square$

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