Counter example for Statement 1:-

$G_{1}=\left ( 2\mathbb{Z},+ \right )$

$G_{2}=\left ( 3\mathbb{Z},+ \right )$

$G_{1}\cup G_{2}=\left ( 2\mathbb{Z}\cup 3\mathbb{Z},+ \right )$

Now as ,$G_{1}\cup G_{2}$ is a group so it must be closed but it isn’t as , $5$ should be in $G_{1}\cup G_{2}$ but it isn’t .