I'm unsure on the following points:
(i) Prove that if a group G has order , where p and q are prime numbers and , then G contains a unique Sylow p-Subgroup. Hence conclude that there are no simple groups of orders 55, 57 or 58.
(ii) Prove that every group of order 59 is simple.
(iii) Prove that there are no simple groups of order 56.
(iv) Give an example of a simple group of order 60, as well as an example of a non-simple group of order 60.
For (i) I know that the number of Sylow p-Subgroups of G divides the order of G and that it has the form ( ). I figured does not divide for (is this right?If so, why?) so there must be only one(unique) Sylow p-Subgroup. , , . If a group is not simple then it has a proper normal subgroup. So groups of these orders have a unique Sylow p-Subgroup and for some reason this means that the Sylow p-Subgroup in question is normal(proper).
For (ii) 59 is a prime so for some reason there is no unique sylow Subgroup.(no subgroup by Lagrange) My question is this: If there is no Sylow Subgroup then is there no subgroup in general?
So therefore there is no proper subgroup of a group of order 59, so no proper normal subgroup, so it's simple?
For (iii) , so there exists a subgroup in this group, say G, of order 2,4,8, and 7. I guess I need to prove that G contains a unique Sylow p-Subgroup but I don't know how.
For (iv) A simple group of order 60 is . Is a non-simple group of order 60 the dihedral group , if so why is this non-simple?
Thanks for any help!