First, closure is not a requirement. Because we define a binary operation such that , so of course it is closed. However, a subgroup needs to satisfy the closure property.

If you define G\times H)\times (G\times H)\mapsto G\times H" alt="*G\times H)\times (G\times H)\mapsto G\times H" /> as where is the binary operation on and is the binary operation on then defines a binary operation.

Proving the condition for to be a group are straightforward. For example, if you chose where is identity element in and where is identity element in . Then . And similarly . Try to prove the other properties.