8.5 Semidirect Products
1. a. Write σ = (1 2) 
Sn and 
Then An ⊆ AnH ⊆ Sn so, since Sn/An ≅ C2, either AnH = Sn or AnH = An. Since σ ∉ An we have Sn = AnH. Similarly, An ∩ H ≠ {ε} means An ∩ H = H (because H is simple), again contradicting h ∉ An. Hence An ∩ H = {ε} and the result follows from Theorem 2.
Sn and Then An ⊆ AnH ⊆ Sn so, since Sn/An ≅ C2, either AnH = Sn or AnH = An. Since σ ∉ An we have Sn = AnH. Similarly, An ∩ H ≠ {ε} means An ∩ H = H (because H is simple), again contradicting h ∉ An. Hence An ∩ H = {ε} and the result follows from Theorem 2.
3. This is an instance of Theorem 3 (3), where p = 3 and q = 13 . We have q ≡ 1 (mod p) so we look for m such that 1 ≤ m ≤ 12 and m3 ≡ 1 (mod 13). If m = 1 then G ≅ C13 × C3 ≅ C55. The first solution with m > 1 is m = 3, whence 
where 
and ab = ba3.
where and ab = ba3.
5. In G, Sylow-3 gives n3 = 1, 10 and n5 = 1, 6 ; if neither is 1 then G has 10 · 2 = 20 elements of order 3, and 6 · 4 = 24 elements of order 5, a contradiction as 
So if P and Q are Sylow 3-and 5-subgroups, then K = PQ is a subgroup of order 3 · 5 = 15 (as P ∩ Q = {1}) . Hence K has index 2, so K G. Moreover both P K and Q K by Sylow-3 ...
So if P and Q are Sylow 3-and 5-subgroups, then K = PQ is a subgroup of order 3 · 5 = 15 (as P ∩ Q = {1}) . Hence K has index 2, so K G. Moreover both P K and Q K by Sylow-3 ...Get Introduction to Abstract Algebra, Solutions Manual, 4th Edition now with the O’Reilly learning platform.
O’Reilly members experience books, live events, courses curated by job role, and more from O’Reilly and nearly 200 top publishers.
