Igor Rapinchuk
1.
Suppose

G

=
p
n
.
Let
S
1
,...,S
k
be the orbits of
G
on
S.
Then

S

=

S
1

+
···
+

S
k

Suppose that
G
does not have a ﬁxed point. Then the size of every orbit is
>
1
.
On the
other hand,

S
i

= [
G
:
G
x
i
] where
x
i
∈
S
i
.
It follows that

S
i

=
p
n
i
for some
n
i
>
0
,
in
particular,

S
i

is divisible by
p.
But then

S

is also divisible by
p.
A contradiction.
2.
Let
G
be the dihedral group of order 2
n,
G
=
{
1
,x,.
..,x
n

1
,y,xy,.
..,x
n

1
y
}
where
x
n
= 1
, y
2
= 1
,
and
yxy

1
=
x

1
.
We have
x
i
x
j
(
x
i
)

1
=
x
j
and (
x
i
y
)
x
j
(
x
i
y
)

1
=
x

j
So, the conjugacy class of
x
j
is
{
x
j
,x

j
}
.
Next,
x
i
(
x
j
y
)(
x
i
)

1
=
x
2
i
(
x
j
y
) and (
x
i
y
)(
x
j
y
)(
x
i
y
)

1
=
x
i
yx
j
yy

1
x

i
=
=
x
2
i

j
y
=
x
2(
i

j
)
(
x
j
y
)
So, the conjugacy class of
x
j
y
is
{
x
2
i
(
x
j
y
)

i
= 0
,...,n

1
}
.
Case 1.
n
is odd,
n
= 2
m
+1
.
Then
x
j
6
=
x

j
for any
j
= 1
,...,n

1
.
So, the powers of
x
split into one 1element conjugacy class
{
1
}
and
m
2element conjugacy classes
{
x
j
,x

j
}
,
j
= 1
,...,m.
Since
n
is odd, any power of
x
can be written in the form
x
2
i
,
so all elements
of the form
x
j
y
form a single conjugacy class. So, the class equation for
G
looks as follows:
2
n
= 1 + 2 +
···
+ 2

{z
}
m
+
n
Case 2.
n
is even,
n
= 2
m.
Then
x
j
=
x

j
iﬀ
j
= 0 or
m,
so there are two 1element
conjugacy classes
{
1
}
and
{
x
m
}
.
The conjugacy class of any other power of
x
contains 2
elements; these classes are
{
x
j
,x

j
}
,
where
j
= 1
,...m

1
.
Since
n
is even, there are
m
elements of the form
x
2
i
.
So, the conjugacy class of an element of the form
x
j
y
contains
m
elements, and therefore there are 2 such classes. So, the class equation for
G
looks as
follows:
2
n
= 1 + 1 + 2 +
...
+ 2

{z
}
m

1
+
m
+
m
3.
Suppose
G/Z
is generated by a coset
gZ.
This means that any coset
xZ
is of the form
g
i
Z
for some integer
i.
It follows that any
x
∈
G
can be written in the form
g
i
z
for
z
∈
Z.
Now, let us take two arbitrary
a,b
∈
G
and write then in the form
a
=
g
i
z
1
and
b
=
g
j
z
2
.
Then
ab
= (
g
i
z
1
)(
g
j
z
2
) =
g
i
+
j
z
1
z
2
and
ba
= (
g
j
z
2
)(
g
i
z
1
) =
g
i
+
j
z
2
z
1
because
z
1
and
z
2
commute with
g.
Since
z
1
z
2
=
z
2
z
1
,
we see that
ab
=
ba,
so
G
is abelian.
1