Solution
= 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 $\langle g\rangle=G$. Thus \b[the group is cyclic, and hence abelian]. In particular, every pair of elements is $g^i,g^j$, and $g^ig^j=g^{i+j}=g^{j+i}=g^jg^i$.