I wanted to find group $A\times B$ such that it has subgroup which is not of form $C\times D$ where $C < A$ and $D <B$.
Actually I think this is wrong I tried to get proof. But my friend told this is true . that it possible to have group with such prpoerty
ANy suggestion is appreciated
Consider $G=\Bbb{Z}_2 \times \Bbb{Z}_2$ and let $H=\langle (1,1) \rangle=\{(0,0), (1,1)\}$. Then $H \leq G$ but $H$ cannot be expressed as $C \times D$. Because if it could be, then $0,1 \in C$ and $0,1 \in D$. This will mean $C=\Bbb{Z}_2$ and same for $D$. But then $C \times D$ will have $4$ elements. Thus cannot be same as $H$.