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?

  • $\begingroup$ You can deduce the existence of an order 14 subgroup from the normality of the Sylow 7-subgroup $P$ and the fact that $G/P$ has order 4, and hence has a subgroup of order 2. $\endgroup$ – Derek Holt Nov 13 '12 at 8:32

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?

  • $\begingroup$ ok now I've managed to show there exists exactly 1 subgroup of order 14, and it contains all 6 elements of order 14 in G. So that should leave exactly 1 element of order 2? Also the class equation is impossible to work out... $\endgroup$ – Benjamin Lu Nov 13 '12 at 8:46
  • $\begingroup$ Is the order of conjugacy classes related to the order of its elements here somehow? $\endgroup$ – Benjamin Lu Nov 13 '12 at 8:49
  • $\begingroup$ Well you should know from Sylow's theorems that the Sylow 2-subgroups are all conjugate to each other, and that should get you a handle on the elements they contain. Then you should be able to take an element of order 4 and identify its stabiliser under conjugation by elements of the whole group - start with the ones you know most about. $\endgroup$ – Mark Bennet Nov 13 '12 at 12:12
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    $\begingroup$ The order of the class of $g$ is the index in $G$ of the order of the centralizer of $g$ in $G$, and you have enough information now to work out centralizer orders. (Note that elements of order 4 and 7 cannot centralize each other, or else the group would be abelian.) $\endgroup$ – Derek Holt Nov 13 '12 at 12:14
  • $\begingroup$ @DerekHolt Ok that should go somewhere... how about the uniqueness of such group? $\endgroup$ – Benjamin Lu Nov 13 '12 at 17:56

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