 Solution

In the vector space ${\mathbb{F}}_2^n$, addition and subtraction agree. Put $V=⟨Y⟩$, the vector subspace spanned by $Y$, andPart b gives $4A\subseteq 2A+V=H$. Conversely $2A\subseteq 4A$, because two copies of any fixed element of $A$ sum to zero. Henceand $0\in H$, so $H$ is a subgroup. For any $a_0\in A$, every $a\in A$ satisfies $a-a_0=a+a_0\in H$, and therefore $A\subseteq a_0+H$.

Because $y_1=0$, the dimension of a vector space $V$ is at most $|Y|-1\le 2K^2-2$. Consequentlywhere the last inequality uses $K\ge 1$. This is the requested bound.