Let $G, H$ be finite groups. Then any subgroup of $G × H$ is equal to $A × B$ for some subgroups $A<G$ and $B<H.$

Question

Let $G, H$ be finite groups. Then any subgroup of $G × H$ is equal to $A × B$ for some subgroups $A<G$ and $B<H.$

Answer is given as False.

1. What does $A<G $means? Does it mean A is a subgroup or proper subgroup?

2.If it was proper subgroup MY ATTEMPT:$G × H$ is a subgroup of itself which is not equal to $A × B$ for all $A<G$ and $B<H.$

3.What if A is not a proper subgroup of G.Does it still false? If yes please give an (counter)example

Answer 1

Let's make the quantifiers in the statement a bit more precise:

For all finite groups $G$ and $H$, and for all subgroups $K$ of $G \times H$, there exist subgroups $A$ of $G$ and $B$ of $H$ such that $K = A \times B$.

To show this statement is false, prove its negative:

There exist finite groups $G$ and $H$, and a subgroup $K$ of $G \times H$, such that for all subgroups $A$ of $G$ and $B$ of $H$, $K \neq A \times B$.

You mentioned the subgroup $K = G\times H$ itself. Notice that this subgroup is in the form $A \times B$, where $A$ is a subgroup of $G$ and $B$ is a subgroup of $H$. Just take $A = G$ and $B=H$. So this doesn't lead to a counterexample.

If you plot $G$ as a set of dots on the $x$-axis, and $H$ as a set of dots on the $y$-axis, then $G \times H$ is a grid of dots in the $xy$-plane. Subgroups of the form $A \times B$ will have a “rectangular” appearance to them, since they contain every single point $(a,b)$, where $a \in A$ and $b \in B$. So try to think about a subgroup that looks more “diagonal”.

Answer 2

A classical counterexample: let $G$ be a non-trivial group and define $D=\{(g,g): g \in G \}$. Then $D$ is a proper subgroup of $G \times G$, but certainly not a direct product of subgroups of $G$.