The Simple Group 168 - Sylow Theory - Part 2.

Note: Part 5 goes off the rails; I can't just assume the subgroup we choose normalizes H_2 a priori. We can still fix with elementary methods and the occasional lucky break.

Fix for Part 5 (2:15) - disregard table: Key to note is that there are no elements of orders 6, 14, or 21 (since |N(H_7)|=21, nonabelian). Thus all elements have order 3, 7, or a divisor of 8. There must be 64 elements of the latter type. (64=168-56-48)

Note that n_2 = 1, 3, 7, or 21.

H_2 is nonabelian.
Proof by contradiction: Suppose H_2 is abelian.
Let x be nonidentity in H_2. We show x belongs uniquely to H_2 among Sylow 2-subgroups. First, this clear if x has order 8. Suppose x has order 2 or 4. If N(x), the normalizer of the subgroup generated by x, has more than 8 elements, the possible orders are 24, 56, or 168. (Key: since H_2 abelian, H_2 is contained in N(x).)

Reject 168 by simple. Next, |N(e, x)|=24 cannot occur: the normal Z/3 subgroup acting on x produces an element of order 6 (x has order 2) or an element of order 12 (x has order 4, Aut(Z/4)=Z/2 - no order three element). Finally, for 56, either an element of order 14 (normal H_7), or N(x) contains a single Sylow 2-subgroup H_2 (48 elements of order 7).

Now H_2 contains 7 nonidentity elements, each unique to H_2 among Sylow 2-subgroups. Under conjugation, these generate either 7, 21, 49, or 147 distinct elements in G, contradiction. Thus the Sylow subgroup H_2 is nonabelian. QED

Extra: Suppose x has order 7. Then |Z(x)|=7, since no elements of order 14, 21. That is, |C_x|=24.

See this MSE post for reference:

https://math.stackexchange.com..../questions/3312332/s

Abstract Algebra: Let G be a simple group of order 168. In Part 2, we compute the number of Sylow-2 subgroups and show that each Sylow-2 subgroup is isomorphic to D_8, the symmetry group of the square.