Math 150a: Modern Algebra
Homework 5 Solutions
2.4.19:
Prove that if a group contains exactly one element of order 2, then that element is in the center of
the group.
Solution:
The key fact here is that conjugation preserves order:
g
n
=
e
⇔
xg
n
x
−
1
=
xex
−
1
=
e
, and
xg
n
x
−
1
= (
xgx
−
1
)
n
.
Let G be a group, and
g
∈
G
be the unique element of order 2. Then,
∀
x
∈
G
the conjugate
xgx
−
1
must also
have order 2. Since
g
is the unique such element,
xgx
−
1
=
g
. Or, multiplying by
x
on the right,
xg
=
gx
.
This exactly means that
g
∈
Z
(
G
)
.
square
2.5.3:
Determine the number of equivalence relations on a set of five elements.
Solution:
Every equivalence relation on a set determines a partition of the set, and vice versa. So we will
count the number of partitions of a set of five elements. Start by considering the largest equivalence class of
the partition. If it only contains one element, then there is only one possible partition to count,
{
}
,
{
}
,
{
}
,
{
}
,
{
}
.
Next, consider the largest equivalence class to be of size 2. There are
(
5
2
)
ways to pick a set with 2 elements.
Then, for each, we may split up the remaining 3 elements as follows:
{
,
} | {
}
,
{
}
,
{
}
[in one way]; and
{
,
} | {
,
}
,
{
}
[in
parenleftbigg
3
2
parenrightbigg
ways];
Counting in this way, we have kept track of the ordering of the two partitions of order 2. Since we don’t
care about this ordering, we should divide by the number of permutations of the ordering. So there are
(
5
2
)(
3
2
)
2
partitions of the form
{
,
}
,
{
,
}
,
{
}
.
To avoid double counting, we next consider the largest equivalence class to be of size 3. Then there are
(
5
3
)
ways to start, with two choices for each:
{
,
,
}
,
{
}
,
{
}
,
and
{
,
,
}
,
{
,
}
.
Finally, there are
(
5
4
)
ways to choose a partition with an equivalence class with 4 elements:
{
,
,
,
}
,
{
}
.
And one partition with an equivalence class with 5 elements:
{
,
,
,
,
}
.
So, all together, we have counted
1
+
parenleftbigg
5
2
parenrightbigg
(
1
+
(
3
2
)
2
)+
parenleftbigg
5
3
parenrightbigg
2
+
parenleftbigg
5
4
parenrightbigg
+
1
=
52
.