Algebra , Assignment 4, Solutions
1. In this problem, assume (important!) that
G
is an
Abelian
. Set
H
=
{
g
∈
G
:
g
5
=
e
}
.
(Warning: Expressions such as
x
1
/
5
are
not
well
defined. Do not use them!)
(a) Show
H
is a subgroup of
G
. Point out where the assumption that
G
was Abelian was used.
Solution. As usual, three parts.
Identity:
As
e
5
=
e
,
e
∈
H
.
Product:
If
x, y
∈
H
then
x
5
=
y
5
=
e
so that (
xy
)
5
=
x
5
y
5
=
ee
=
e
so that
xy
∈
H
– where we need that the group is Abelian
to say that
(
xy
)
5
=
xyxyxyxyxy
=
xxxxxyyyyy
=
x
5
y
5
Inverse:
If
xinH
then
x
5
=
e
so that (
x

1
)
5
= (
x
5
)

1
=
e

1
=
e
so
x

1
∈
H
.
(b) Show
H
is a normal subgroup of
G
.
G
is Abelian so that all subgroups of
G
are Normal!
(c) Assume further that
G
is finite and that
H
=
{
e
}
.
Show that
the map
φ
:
G
→
G
given by
φ
(
g
) =
g
5
is an automorphism.
(Definition: An automorphism is an isomorphism from a group
to itself.)
Proof:
φ
is a homomorphism as
φ
(
xy
) = (
xy
)
5
=
x
5
y
5
=
φ
(
x
)
φ
(
y
) for all
x, y
∈
G
φ
is injective as the kernal
K
φ
=
{
g
∈
G
:
φ
(
g
) =
e
}
=
H
=
{
e
}
by assumption, and we proved in class that
φ
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 Fall '08
 JOSEPHS
 Algebra, Normal subgroup, Inverse function, Abelian

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