Partial Solution Sketches, Math 74 HW7
(7.1a) Not associative:
a

(
b

c
) =
a

b
+
c
= (
a

b
)

c
.
(7.1b) No inverse: 2

1
= 1
/
2
/
∈
Z
.
(7.1c) Group. (Check that the axioms hold. The most important one is closure under addition – how do
you know
a/b
+
c/d
has an odd denominator if
b, d
are odd?)
(7.3) Let
a, b
∈
G
.
(
ab
)
2
=
a
2
b
2
(
ab
)(
ab
) =
abab
abab
=
aabb
groups are associative
a

1
ababb

1
=
a

1
aabb

1
group elements have inverses
ba
=
ab
Hence
G
is abelian.
(7.4) Since a group must have an identity, we may write
G
=
{
e, x, y
}
.
Trivially,
x
·
e
=
e
·
x
and
y
·
e
=
e
·
y
. Moreover, since the identity is unique,
xy
=
x
(as
y
=
e
) and
xy
=
y
(as
x
=
e
). Hence
xy
=
e
.
By the same argument,
yx
=
e
, so
xy
=
yx
as desired.
(7.5) Let
G
=
{
a
1
, . . . , a
n
}
be a group of order
n
.
For
a
i
∈
G
, consider
a
i
, a
2
i
, a
3
i
, . . . , a
n
+1
i
, which are
n
+ 1 elements among the
n
elements of
G
by closure of groups. By the pigeonhole principle, there exist
t, s
such that
a
t
i
=
a
s
i
and
t > s
. Then
a
t

s
i
=
e
; define
N
i
=
t

s
. Now consider
N
=
n
i
N
i
. For each
a
i
∈
G
, we then have
a
N
i
=
a
N
1
···
N
i
···
N
n
i
=
a
N
i
·
j
=
i
N
j
i
= (
a
N
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 Spring '09
 DANBERWICK
 Addition, Division, Inverses, 1 g, Φg, Partial Solution Sketches, G. Let 2Z

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