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.
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.
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
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.
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.
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.
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.