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Exam 1 Solutions
1. (a) Not a group. 0 has no inverse.
(b) Not a group. It’s not associative.
x/
(
y/z
) =
xz/y
, (
x/y
)
/z
=
x/yz
.
(c) Is a group.
•
Closure
: Let
x, y,
∈
G
.
x
*
y
=
x
+
y
+
xy
=

1
⇔
(
x
+ 1)(
y
+ 1) = 0
⇔
x
=

1 or
y
=

1. Since

1
6∈
G
, we have closure.
•
Associative
:
x
*
(
y
*
z
) =
x
*
(
y
+
z
+
yz
) =
x
+
y
+
z
+
yz
+
xy
+
xz
+
xyz
.
(
x
*
y
)
*
z
= (
x
+
y
+
xy
)
*
z
=
x
+
y
+
z
+
xy
+
xz
+
yz
+
xyz
. Thus associativity
holds.
•
Identity
:
x
*
0 =
x
+ 0 + 0 =
x
and 0
*
x
= 0 +
x
+ 0 =
x
. Thus 0 is the identity.
•
Inverses
:
x

1
=

x
1+
x
∈
G
(since

1
6∈
G
).
x
*
(

x
1+
x
) =
x
+

x
1+
x
+

x
2
1+
x
= 0. By
commutativity of addition and multiplication of reals, we get

x
1+
x
*
x
= 0 as well.
2.
e
∈
B
and
f
(
e
) =
e
so
e
∈
f

1
(
B
)
6
= Ø. Let
x, y
∈
f

1
(
B
)
⇒
f
(
x
)
, f
(
y
)
∈
B
. Thus
f
(
x
)
f
(
y
)

1
∈
B
since
B
is a group. Hence
f
(
xy

1
)
∈
B
since
f
is a homomorphism. So
xy

1
∈
f

1
(
B
) and
f

1
(
B
)
≤
G
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This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.
 Spring '11
 Johnson
 Algebra

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