Prove that if such a group exists it is unique (up to isomorphism). Also, determine the numbers of elements of each order, and the class equation of the group.
I don't really know how to do the unique part… A brute force way could be to classify groups of order 28 and look at each isomorphism classes, but I feel like there should be a simpler way.
As for the numbers of elements, I think there are 14 of order 4 because each order 4 subgroups contains 2 elements of order 4 and there are 7 of distinct such groups since its Sylow 2's are all cyclic and can't overlap anywhere else except identity; and there are 6 elements of order 7 since there is only 1 group of order 7 (Sylow 7). There is at most 1 subgroup of order 14, since if there are two then $G$ would be isomorphic to $C_{14} \times C_{14}$. If there is indeed one, then it contains 6 elements of order 14, and that means there is one element of order 2.
But I can't prove there is (or isn't) an order 14 subgroup… I suppose I could use the fact that there are 4 isomorphism classes of groups order 28, and only $D_{28}$ and $D_{14} \times C_2$ are nonabelian and they both contain an order 14 subgroup.
As for the class equation, subgroup order 14 is normal, there are 7 Sylow 2-subgroups which are conjugates of each other, Sylow 7-subgroup is normal so $|G|$ = 28 = 1 + 7 + 1 + 1 +… The rest must sum up to 18 and each term divides 28.
Above is all I've got so far. Please correct me if I made a mistake, and help me with the part I haven't got yet?
Best Answer
It is not automatic that the order 4 subgroups have only the identity in common - they could (in principle) have an element of order 2 in common.
You have 7 subgroups of order 4 and one of order 7 you know there is at least one element of order 2.
I suggest analysing the orders of the elements you already have, knowing that you have identified everything of order 7 and order 4 (and order 1). How many elements are not in your list? What are the options for the orders of elements which you haven't yet pinned down?