Introduction to Abstract Algebra, Solutions Manual, 4th Edition

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; (ba^k)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⁻¹ ⊆ aHa⁻¹ ⊆ H because H G. If h H, we must show h