Romanian Mathematical Olympiad

a) Since $G$ is a group, it follows that $|xH|=p$. Therefore $$n=|G|\ge |H\cup xH|=|H|+|xH|-|H\cap xH|=2p-|H\cap xH|,$$ whence $|H\cap xH|\ge 2p-n$.

b) Let have $x\in H$ and $y\in H\cap xH$. Then $y=y^{-1}$ and $y=xh$, $h\in H$. Since $xy=x^{2}h=h\in H$, it follows that $xy=(xy{)}^{-1}=y^{-1}{x}^{-1}=yx$. Therefore $x$ commutes with all elements of $H\cap xH$. Since $|H\cap xH|\ge 2p-n>\frac{3n}{2}-n=\frac{n}{2}$, it follows the subgroup of the elements that commute with $x$ has at least $\frac{n}{2}$ elements. From Lagrange's theorem, it follows that $x$ commutes with all elements of $G$. Therefore $H\subseteq Z(G)$, the center of $G$. It results $|Z(G)|\ge |H|=p>\frac{3n}{4}>\frac{n}{2}$, so $Z(G)=G$.

c) Assume $G$ is commutative. If $x,y\in H$, then $(xy{)}^2=x^2{y}^2=e$, so $xy\in H$. Since $H$ is a finite set, it follows that $H$ is a subgroup of $G$. The condition $|H|>\frac{n}{2}$ forces $H=G$, so $\frac{3n}{4}\ge |H|=|G|=n$, contradiction.