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 applied to K, whence K ≅ P × Q ≅ C3 × C5 ≅ C15, say 
where 
Let H be any Sylow 2-subgroup, say 
where 
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 applied to K, whence K ≅ P × Q ≅ C3 × C5 ≅ C15, say where Let H be any Sylow 2-subgroup, say where
Let b−1ab = am, so 
for k ≥ 1 . Since b2 = 1 this gives 
that is m2 ≡ 1 (mod 15). The solutions are m = ± 1 and m = ± 4.
for k ≥ 1 . Since b2 = 1 this gives that is m2 ≡ 1 (mod 15). The solutions are m = ± 1 and m = ± 4.
Case 1. m = 1 . Then ab = ba so G is abelian, and we get G ≅ C15 × C3 ≅ C30.
Case 2. m = − 1 . Then ab = ba−1, so aba = b and G ≅ D15.
Case 3. m = 4 . Then b−1ab = a4. Take a1 = a3 so 
Then 
It follows that 
Define 
Then U ∩ V = {1} because V6 ⊆ U. It follows that UV = G. Finally, one verifies that a5b = ba5, so U and V commute elementwise. This gives G ≅ U × V ≅ D5 × C3.
Then It follows that Define Then U ∩ V = {1} because V6 ⊆ U. It follows that UV = G. Finally, one verifies that a5b = ba5, so U and V commute elementwise. This gives G ≅ U × V ≅ D5 × C3.
Case 4. m = − 4 . Here b−1a5b = (b−1ab)5 = a−20 = (a5)−1. Hence 
Take 
so that U ∩ V = {1} . Hence UV = G. Moreover, since b−1a3b = (b−1ab)3 = a−12 = a3, U and V commute elementwise, so G ≅ U × V ≅ D3 × C5.
Take so that U ∩ V = {1} . Hence UV = G. Moreover, since b−1a3b = (b−1ab)3 = a−12 = a3, U and V commute elementwise, so G ≅ U × V ≅ D3 × C5.
Finally, observe that no two of the groups C30, D15, D5 × C3 and D3 × C5 are isomorphic: C30 is the only abelian one; D15 and D3 × C5 have no element of order 6, while D5 × C3 has 10; D15 has 15 elements of order 2, while D3 × C5 has only 3.
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