Solution. Since (ab)(abc) = (bc) and (abc)(ab) = (ac), it is easy to see that the center of S3 is the trivial subgroup. Therefore, the group of inner automorphisms of S3 is isomorphic to S3 and has size 6. On the other hand, S3 has 3 transpositions. These must be permuted by an automorphism of S3, and a nontrivial automorphism induces a nontrivial permutation. This gives an injective homomorphi...

Solution. Let np denote the number of p-Sylow subgroups. Since n11 is 1 mod 11 and divides 12, we know n11 is either 1 or 12. If it is 1, then it is normal and we are done. So instead, assume n11 is 12. Then there must be 120 elements of order 11 in G, leaving only 12 more elements available. Since n3 is 1 mod 3 and divides 44, we know that n3 is 1, 4 or 22. If it is 1, then again it is normal ...

Solution. The prime factorization of 185 is 5*37. Given a group of order 185, Let n5 be the number of subgroups of order 5 and n37 the number of subgroups of order 37. Since subgroups of order 5 and 37 will be Sylow subgroups, by the Sylow Theorems n5 and n37 have to be 1 (mod 5) and 1 (mod 37), respectively and divide 37 and 5, respectively. Hence, n5 = n37 = 1 and the 5 and 37 Sylow subgroups...

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