Math 150a: Modern Algebra
Homework 6 Solutions
GK1.
Show that if
G
is a finite group and
p
is a prime number, then the number of elements of order
p
in
G
is divisible by
p

1. The result is certainly not true if
p
is not prime; be careful to explain where
the argument uses this assumption. (Hint: This is a generalization of a midterm problem. Look at the
solution to that problem and first try the case
p
=
5.)
Solution:
If
G
is a finite group, then there are a finite number of elements of order
p
. We will begin to count
them off in groups of
p

1 (which will show that the number of elements of order
p
is divisible by
p

1). First, let’s check that the result still holds true when there are
no
elements of order
p
. Then
0
= (
p

1
)
·
0.
Now, given any one element
g
∈
G
of order
p
, all powers of
g
(except the identity) have order
p
. That
is, since
g
p
=
1; each of
g
1
,
g
2
,
g
3
,...,
g
p

1
also have order
p
. This is because for 1
≤
k
≤
p

1
(
g
k
)
n
=
1
⇒
g
kn
=
1
⇒
p

kn
,
because
p
is the order of
g
⇒
p

k
or
p

n
,
since
p
is prime
⇒
p

n
,
since
p
is prime and 1
≤
k
≤
p

1
.
Thus, the order of
g
k
is
p
since
(
g
k
)
p
= (
g
p
)
k
=
1
k
=
1. [If
p
wasn’t prime, then
g
k
would have order
p
only when
(
k
,
p
) =
1 (
k
relatively prime to
p
).]
To finish the proof, consider the set
{
g
1
,
g
2
,...,
g
p

1
}
. It either contains every element in
G
of order
p
, or it doesn’t. If not, then there is another
h
∈
G
of order
p
that’s not a power of
g
. Then we know
that each of the
p

1 distinct powers of
h
also have order
p
, and the set
{
g
1
,
g
2
,...,
g
p

1
,
h
1
,
h
2
,...,
h
p

1
}
(again) either contains every element of
G
of order
p
or doesn’t. Continuing in this way, we will
eventually count every element of
G
of order
p
, since there are a finite number of elements in
G
.
GK2.
One form of Lagrange’s theorem says that the order of every element
g
∈
G
divides the order of
G
. Cauchy’s theorem is a partial converse, that if a prime
p
divides the order of
G
, then there is an
element of order
p
.
a.
Given two distinct primes
p
and
q
, find a group
G
whose order is divisible by both
p
and
q
, but
which has no element of order
pq
. (Hint: It’s one of the standard groups of the course.)
Solution:
Given primes
p
and
q
, assume
p
<
q
. Consider the symmetric group
S
q
. Then

S
q

=
q
!
=
q
(
q

1
)
···
p
···
2
·
1
,
but
S
q
has no element of order
pq
. Since
p
and
q
are both prime,
any
permutation of order
pq
must either be a
pq
cycle or a product of a
p
cycle and
q
cycle (when written as a product of
disjoint cycles). But
S
q
acts on the numbers 1–
q
, and there are simply not enough numbers for
either possibility.
b.
For each
n
, find a group
G
with
n
elements that has an element of order
a
for every divisor
a

n
.
I.e.
,
it should be the same
G
for each
a
. (Hint: Again, it’s one of the standard groups of the course.)
Solution:
The cyclic group
C
n
=
{
1
,
x
,
x
2
,
x
3
,...,
x
n

1
}
has
n
elements. And for every
a
that divides
n
, we can find a
b
so that
ba
=
n
. Then
(
x
b
)
a
=
x
ba
=
x
n
=
1. And for each
k
<
a
,
bk
<
n
⇒
(
x
b
)
k
=
1. Thus the order of
x
b
is
a
.
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 Spring '03
 Kuperberg
 Algebra

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