# 2.8 Normal Subgroups

1. a. H = {1, a6, b, ba6} is a subgroup because It is not normal in D4 because Ha = {a, a7, ba, ba7} while aH = {a, a7, ba11, ba5}.
c. This is closed because a2b = ba10, a4b = ba8, a6b = ba6; (bak)2 = 1 for each k. Hence H is a subgroup. It is normal, being of index 2.
3. A4 has no subgroup of order 6 [Exercise 34 §2.6] so the only subgroups (normal or not) have order 1, 2, 3 or 4. If , it contains no 3-cycle and so equals K. Thus K A4 being unique of its order. If , let H = (1, 2, 3) without loss of generality. But (1 4)−1(1 2 3)(1 4) = (2 3 4) shows H is not normal. Finally without loss of generality. But (1 3)−1(1 2)(1 3) = (2 3) shows H is not normal.
5. First aKa−1 is a subgroup, by Theorem 5 §2.3, and aKa−1aHa−1H because H G. If h H, we must show h

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