How many elements of order 7 in a simple group of order 168?

Answer 1

Let G be a simple group of order 168. We know the number of Sylow 7 subgroups is congruent to 1 mod 7.(Sylow's 3rd theorem). If N is the number of Sylow 7 subgroups, then N =1,8,15,22,29,... Since this is a Sylow 7 subgroup, we know that each group has exactly 7 elements. Hence, we can eliminate any number greater than 22. Hence the only choices left are 1, 8, 15, and 22. Since we know the number of Sylow 7 subgroups divides 168 = 2^3*3*7, we can eliminate 15 and 22. Thus there can only be either 1 or 8 Sylow 7 subgroups. If there was a unique Sylow 7 subgroup, then again by a corollary to Sylow theory, this group would be normal and thus contradict the simplicity assumption. Hence there are exactly 8 Sylow 7 subgroups.

Now to answer your question: Each Sylow 7 subgroup contains 6 elements of order 7 (the other element is the identity and has order 1). Thus, there should be 8*6 =48 elements of order 7 in any simple group of order 168.

Answer 2

No 2 of the 7 sylow communities can overlap different than on the identity, because of the fact 7 is a main sort. The intersection is a subgroup whose order will could divide 7, and for suited interesection that could in simple terms be a million. what number aspects of order 7 are there? There are 6 for all the 8 sylow subgroups for an entire of 6x8=40 8. What are the different fifty six-40 8 = 8 aspects? they are able to easily be the a million sylow team of order 8 - there are actually not adequate aspects there for there to be greater advantageous than a million such subgroup.

Answer 3

24
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