By the Third Sylow Theorem, there are 1+3k Sylow 3-subgroups, each of order 9, for some k=0,1,2,…. Also, 1+3k must divide 11; hence, there ca...By the Third Sylow Theorem, there are 1+3k Sylow 3-subgroups, each of order 9, for some k=0,1,2,…. Also, 1+3k must divide 11; hence, there can only be a single normal Sylow 3-subgroup H in G. Similarly, there are 1+11k Sylow 11-subgroups and 1+11k must divide 9. Consequently, there is only one Sylow 11-subgroup K in G. By Corollary 14.16, any group of order p2 is abelia…
