Hence, ker(
α
) =
\
P
∈
Syl
3
(
G
)
N
G
(
P
) =
\
P
∈
Syl
3
(
G
)
P
=
{
e
}
.
Now,

α
(
G
)

=

G

= 12 and

S
4

= 24, so

S
4
:
α
(
G
)

= 2. If (
acb
)
∈
α
(
G
),
then (
acb
)
2
= (
abc
)
∈
α
(
G
), so
α
(
G
) has every 3cycle, and the 3cycles
generate
A
4
. Hence,
α
(
G
) =
A
4
.
Thus,
G
=
G/
ker(
α
)
’
A
4
, by the First Isomorphism Theorem.
3.
Let
G
be a group with

G

= 30
. Show that
G
is not simple.
Solution.
We have,

G

= 30 = 2
·
3
·
5. So, from Sylow’s Third Theorem
and the corollary to Sylow’s Second Theorem, we know

Syl
5
(
G
)

divides
6 and

Syl
5
(
G
)
 ≡
1(mod 5), so that either

Syl
5
(
G
)

= 1 or

Syl
5
(
G
)

= 6.
Similarly,

Syl
3
(
G
)

divides 10 and

Syl
3
(
G
)
 ≡
1(mod 3), so that either

Syl
3
(
G
)

= 1 or

Syl
5
(
G
)

= 10.
If

Syl
3
(
G
)

= 10 and

Syl
5
(
G
)

= 6, then there are 20 nontrivial elements
of order 3 and 24 nontrivial elements of order 5, which is a contradiction
because

G

= 30.
So,

Syl
p
(
G
)

= 1 for at least one value of
p
∈ {
3
,
5
}
.
But if

Syl
p
(
G
)

= 1, then by the corollary to Sylow’s Second Theorem,
P
C
G
for some
P
∈
Syl
p
(
G
)

, and hence
G
is definitionally not simple.
So, if

G

= 30, then
G
is not simple.
4.
Inside
GL
n
(
C
)
, consider the subgroup
B
consisting of all upper triangular
invertible matrices. Compute the derived series of
B
.
Solution.
Consider
B
≤
GL
n
(
C
) and choose
X, Y
∈
B
.
Let
A
=
XY X

1
Y

1
.
Then, we have,
det(
A
) = det(
XY X

1
Y

1
)
= det(
X
)det(
Y
)det(
X

1
)(
Y

1
)
= 1
.
But det(
A
) =
n
Y
i
=1
a
ii
, so
A
only has 1s on its main diagonal.
Hence, the first derived subgroup
B
(1)
is generated by matrices of the form
1
*
. . .
*
0
1
*
.
.
.
.
.
.
.
.
.
.
.
.
*
0
0
. . .
1
.
2