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Math 541
Solutions to HW #8
The following are from Gallian, Chapters 7 and 9. The numbering is ﬁrst given for the 7th edition, and
is then followed by the 6th edition’s numbering (if the numbering is diﬀerent).
•
# 7.15
: Suppose

G

=
pq
with
p
and
q
prime numbers. By Lagrange’s Theorem, the possible
subgroups of
G
have orders 1,
p
,
q
, or
pq
. For a subgroup
H
of
G
to be proper, it must have order
less than
G
, so we can immediately eliminate
pq
. In the case that

H

= 1,
H
contains only
e
so it
is clearly cyclic. In the case that

H

=
p
or

H

=
q
, then
H
has prime order, so it must be cyclic.
This completes the argument.
•
# 7.20
: Since
H
and
K
are both subgroups of
G
, we know that
H
∩
K
must contain
e
, so it is
nonempty. Suppose there is some other element,
g
, that is in
H
∩
K
. Then
g
∈
H
and
g
∈
K
, so

g

>
1 must divide the order of
H
and of
K
, since both
H
and
K
are themselves groups. But, since

H

= 12 and

K

= 35 have no common divisors greater than 1, we have derived a contradiction.
Thus, the only element in
H
∩
K
is
e
, so

H
∩
K

= 1.
•
# 7.39
(resp.
# 7.35
): Since
G
has elements of orders 1 through 10, it’s clear that if

G

=
k
, then
10

k
, 9

k
, ..., 2

k
. Thus, the least possible order of
G
is
lcm
(1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10) which is
2
3
·
3
2
·
5
·
7 = 2520.
•
# 9.1
: No,
H
is not normal in
S
3
. Consider that (13)
H
=
{
(13)
,
(123)
} 6
=
H
(13) =
{
(13)
,
(132)
}
.
•
# 9.7
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This note was uploaded on 11/03/2010 for the course MATH 541 taught by Professor Pollack during the Fall '09 term at BU.
 Fall '09
 Pollack
 Math, Algebra

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