Mathematics 113: Homework # 5
Curtis Tekell
Dr. Poirier
October 11, 2010
Exercise 1.
Let
G
be a group and let
S
be a set with a binary operation
×
. Let
f
:
G
→
S
be a surjective map of sets such that
f
(
ab
) =
f
(
a
)
×
f
(
b
)
for all
a, b
∈
G
. Then
S
is a group
under
×
and
f
is a group homomorphism.
Proof.
Let
a, b, c
∈
S
, and suppose
x, y, z
∈
G
with
f
(
x
) =
a
,
f
(
y
) =
b
and
f
(
z
) =
c
. As the
operation on
G
is associative, we wee that
[
a
×
b
]
×
c
= [
f
(
x
)
×
f
(
y
)]
×
f
(
z
) =
f
(
x
)
×
[
f
(
y
)
×
f
(
z
)] =
a
×
[
b
×
c
]
,
and
×
is associative. Let 1
S
=
f
(1
G
). Then 1
S
a
=
f
(1
G
x
) =
f
(
x
) =
a
=
f
(
x
) =
f
(
x
1
G
) =
a
1
S
, and 1
S
is an identity for
S
. Additionally,
af
(
x

1
) =
f
(
xx

1
) =
f
(1
G
) = 1
S
=
f
(1
G
) =
f
(
x

1
x
) =
f
(
x

1
)
a
and
a

1
=
f
(
x

1
), thus all elements of
S
are invertable. It follows that
S
is a group.
Exercise 2.
The canonical epimorphism
π
:
G
→
G/H
is in fact an epimorphism with
Ker
(
π
) =
H
Proof.
Given
gH
∈
G/H
,
π
(
g
) =
gH
and
π
is surjective.
π
(
ab
) =
abH
=
aHbH
=
π
(
a
)
π
(
b
)
shows that
π
is a homomorphism. Finally,
Ker
(
π
) =
{
g
∈
G
:
gH
=
H
}
=
H
.
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 Summer '11
 gelfand
 Chemistry, Konrad Zuse, Epimorphism, Dr. Poirier, Curtis Tekell

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