Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 9 / b / Solution

Choose $g\ne e$. By Lagrange's theorem, its cyclic subgroup has order dividing the prime 101. It is not of order one, so $⟨g⟩=G$. Thus the group is cyclic, and hence abelian. In particular, every pair of elements is $g^i,g^j$, and $g^i{g}^j=g^{i+j}=g^{j+i}=g^j{g}^i$.