Math 541
Solutions to HW #6
The following are from Gallian, Chapters 4 and 5 (6th edition).
•
# 4.8
: Let
a
be an element of a group and let

a

= 15. Compute the orders of the following elements
of
G
:
–
a
3
,
a
6
,
a
9
,
a
12
*
For each
a
k
above, gcd(15
, k
) = 3. Thus, the order of each is 15
/
3 = 5.
–
a
5
,
a
10
*
For each
a
k
above, gcd(15
, k
) = 5. Thus, the order of each is 15
/
5 = 3.
–
a
2
,
a
4
,
a
8
,
a
14
*
For each
a
k
above, gcd(15
, k
) = 1. Thus, the order of each is 15
/
1 = 15.
•
# 4.14
: Suppose that a cyclic group
G
has exactly three subgroups:
G
itself,
{
e
}
, and a subgroup of
order 7. What is

G

? What can you say if 7 is replaced with
p
where
p
is a prime?
–
Since
G
is cyclic, there is some element
a
in
G
such that
h
a
i
=
G
. Since
G
has a subgroup of
order 7, and
G
is cyclic, we know that 7 divides the order of
G
. That is,
 h
a
i 
=

G

= 7
n
for
some positive integer
n
. We now test a few possible values of
n
:
*
Suppose
n
= 1. Then
G
and one of its subgroups both have order 7. By the Fundamental
Theorem of Cyclic Groups (FTCG),
G
and its subgroup of order 7 are the same, contradicting
the condition that
G
has 3 distinct subgroups.
*
Suppose
n
is 2, 3, 4, 5, or 6. Then, by FTCG,
G
=
h
a
i
has a subgroup of order
n
. Thus,
G
has at least 4 subgroups:
{
e
}
, the subgroup of order 7, the subgroup of order
n
, and
G
itself.
This contradicts the fact that
G
has exactly three subgroups.
*
Suppose
n
= 7. Then

G

= 7
·
7 = 49. Since 7 is the only positive divisor of 49 between 1
and 49, it is the only possible order of a subgroup other than
{
e
}
or
G
. FTCG also tells us
that there is
exactly
one subgroup of order 7. This fits the supposed criteria.
*
In general, if we suppose that
n
is any positive integer besides 7, we see that
G
is guaranteed a
subgroup of order
n
by the FTCG, which means that
G
will have
at least
4 distinct subgroups.
We therefore conclude that the order of
G
must be 7
2
= 49.
–
More generally, if 7 is replaced by any prime
p
under the supposed conditions, the the order of
G
must be
p
2
.
•
# 4.16
: Find a collection of distinct subgroups
h
a
1
i
,
h
a
2
i
, ...,
h
a
n
i
of
Z
240
with the property that
h
a
1
i ⊂ h
a
2
i ⊂
...
⊂ h
a
n
i
with
n
as large as possible.
–
Since
Z
240
is cyclic and the order of a subgroup of a cyclic group divides the order of the group
in which it is contained, we see it must be true that
 h
a
i
i 
divides
 h
a
i
+1
i 
,...,
 h
a
n

1
i 
,
 h
a
n
i 
.
That is, the order of a subgroup divides the order of
every
subgroup in which it is contained.
–
Breaking 240 into its prime factorization, we get 240 = 2
4
·
3
·
5. That is, 240 is the product of 6
primes (note that they need not be distinct).
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 Fall '09
 Pollack
 Math, Algebra, Divisor, Subgroup, Cyclic group, FTCG

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