Why must $a$ be in one of these two cosets: $H, Ha$? [duplicate]

HINT: Note that $H$ and $Ha$ cannot intersect, otherwise there would be a $h,h' \in H$ such that $h' = ha$, which would imply $h^{-1}h'=a$, which would imply $a \in H$.

Remember that the cosets are exactly the equivalence classes of the equivalence relation $a\sim b\iff b^{-1}a\in H$. We know that if we define an equivalence relation on a set $X$, the disjoint union of the equivalence classes is exactly $X$. So $G=H\cup aH$. Since $a^2\in G$, it must be in one of them.

$H$ has index $2=600/300$ in $G$, hence it is normal in $G$ and $G/H$ is isomorphic to the group of order $2$, hence for every $a\in G$, $a^2H= H$, so $a^2\in H$.