75th Romanian Mathematical Olympiad

a) Denoting by $e$ the unit element of the group $G$, for any $x\in G\setminus H$ we have $x^{-1}\in G\setminus H$, so that $e=f(e)=f(x\cdot x^{-1})=g(x)\cdot h(x^{-1})=g(x)\cdot h(x{)}^{-1}$, and it follows that $g(x)=h(x)$ for any $x\in G\setminus H$. For any $a\in H$, choosing an $x\in G\setminus H$ we have that $ax,x^{-1}\in G\setminus H$, so: $$g(a)=g(ax\cdot x^{-1})=g(ax)\cdot g(x^{-1})=h(ax)\cdot h(x^{-1})=h(ax\cdot x^{-1})=h(a).$$ We conclude that $g=h$.

b) Let $G$ be nonabelian and $H=Z(G)$. According to part a), we have $g=h$, so that the relation in the statement becomes

For any $a\in Z(G)$ and $x\in G\setminus Z(G)$, we have $ax,x^{-1}\in G\setminus Z(G)$, so: $$f(a)=f(ax\cdot x^{-1})=g(ax\cdot x^{-1})=g(a).$$ Consider $x\in G\setminus Z(G)$. Then there is an element $y\in G\setminus Z(G)$ such that $xy\ne yx$. If $xy\in Z(G)$ then $$xy=y(xy)y^{-1}=yx\ne xy,$$ which would be a contradiction. Consequently, $xy\in G\setminus Z(G)$, and, since $y^{-1}\in G\setminus Z(G)$, we have: $$f(x)=f(xy\cdot y^{-1})=g(xy\cdot y^{-1})=g(x).$$ We conclude that $f=g$, hence $f=g=h$.