1. The problem statement, all variables and given/known data Assume that G is a group of order 63 that has two Sylow subgroups whose intersecion is non-trivial. Show that G has an element of order 21. 2. Relevant equations 3. The attempt at a solution So by Sylow's theroem, I know that 63=3^2*7, that the sylow 7 subgroup is a normal subgroup of G with 6 order 7 elements. My sylow 3 subgroup is of order 9 and there can be one or 7 sylow three subgroups. Since I know that s7 (sylow 7 subgroup) is normal. So since we have two sylow subgroups whose intersection is non-trival, ths means that we have 7 sylow 3 subgroups. This means we have 14 elements of order 3 in are group G. Since we have a unique sylow 7 subgroup, we have 6 elements of order 7 in G. Let H = sylow 7 subgroup. So for every a in G, aha^-1 is a subgroup of order 7, so we must have aHa^-1=H (since H is normal). So N(H) = G (N(H)= normalizer of H). SInce H has prime order, it is cyclic and abelian. So C(H) contains H (C(H)= centralizer of H in G). So 7 divides |C(H)| and |C(H)| divides 63. So C(H)= H or C(H)=G, or |C(H)|= 21. Is this enough to get that xy=yx wth |x,y|=21?