Find a counterexample Assume that $(G,*)$ and $(H$,#$)$ are groups, and define $@$ on $G \times H$ by if $(g,h) \in G \times H$, and $(a,b) \in G \times H$, then $(g,h)@(a,b) =(g*a, h$#$b).$
(a) Assume that $(A,* )$ is a subgroup of $(G,*)$ and $(B,$ #$)$ is a subgroup of $(H,$ #$)$.
Determine if $A\times B$ must be a subgroup of $G
\times H$?
(b) Assume that $D$ is a subgroup of $G\times H$. Determine if there must exist subgroups $A$ of $G$ and $B$ of $H$ such that $D=A\times B$
 A: For the counter example take $\mathbb{Z}/2\times \mathbb{Z}/2$ and the operation in each group is addition (it is done component wise). Then note that $D=\{(0,0),(1,1)\}$ is indeed a subgroup (check it). The only subgroups of $\mathbb{Z}/2$ are the trivial subgroup or the entire group. Check that neither combination of subgroups will give you $D$. 
Added: $\mathbb{Z}/2$ are the integers modulo $2$. You will learn that later. Think of them right now as the group that has only two elements: $0$ and $1$, and you have $0+0=0$, $0+1=1$, $1+1=0$. Check that this is indeed a group and the identity of the group is $0$. Then consider the group that is given by the product of two copies of this group. Hence, we have elements $(0,0),(1,0),(0,1),(1,1)$, and the operation in this new group is done "component wise", meaning for example: $(1,0)+(1,1)=(1+1,0+1)=(0,1)$. 
The subgroups of a group with $2$ elements are either the trivial subgroup (the subgroup that contains only the identity) or the entire group. Let us denote this two groups by $A$ and $\mathbb{Z}/2$. Check all the possible combinations: 
$A\times A$, $A\times \mathbb{Z}/2$, $\mathbb{Z}/2\times A$ and $\mathbb{Z}/2\times \mathbb{Z}/2$. Compute them and see that none of then give you the $D$ I wrote above. 
