Arkansas Tech University
MATH 4033: Elementary Modern Algebra
Dr. Marcel B. Finan
22
Quotient Groups
Let’s look closely on the construction of the group (ZZ
n
,
⊕
)
.
We know that the
elements in ZZ
n
are equivalence classes of the equivalence relation defined on ZZ
by
a
∼
b
if and only if
n

(
a

b
)
.
Also, an element of ZZ
n
is a right coset. Indeed, if 0
≤
k < n
then
< n >
+
k
= [
k
] =
{
nq
+
k
:
q
∈
ZZ
}
.
The operation
⊕
is defined by
[
a
]
⊕
[
b
] = [
a
+
b
]
or using right cosets
(
< n >
+
a
)
⊕
(
< n >
+
b
) =
< n >
+(
a
+
b
)
.
Note that the cyclic group
< n >
is a normal subgroup of ZZ since ZZ is Abelian
(See Example 21.4). We are going to use the above ideas to construct new groups
where
G
is a group replacing ZZ and
N
is a normal subgroup of
G
playing the
role of
< n > .
More precisely, we have the following theorem.
Theorem 22.1
Let
N
be a normal subgroup of a group
G
and let
G/N
be the set of all right
cosets of
N
in
G.
Define the operation on
G/N
×
G/N
by
(
Na
)(
Nb
) =
N
(
ab
)
.
Then
(
G/N,
·
)
is a group, called the
quotient group
of
G
by
N.
Proof.
·
is a welldefined operation
We must show that if (
Na
1
, Nb
1
) = (
Na
2
, Nb
2
) then
N
(
a
1
b
1
) =
N
(
a
2
b
2
)
.
To
see this, since (
Na
1
, Nb
1
) = (
Na
2
, Nb
2
) then
Na
1
=
Na
2
and
Nb
1
=
Nb
2
.
Since
a
1
=
ea
1
∈
Na
1
then
a
1
∈
Na
2
so that
a
1
=
na
2
for some
n
∈
N.
Sim
ilarly,
b
1
=
n
0
b
2
for some
n
0
∈
N.
Therefore,
a
1
b
1
=
na
2
n
0
b
2
.
Since
N / G
then
a
2
n
0
a

1
2
∈
N,
say
a
2
n
0
a

1
2
=
n
00
∈
N.
Hence,
a
2
n
0
=
n
00
a
2
so that
a
1
b
1
=
nn
00
a
2
b
2
.
But
nn
00
∈
N
so that
a
1
b
1
∈
N
(
a
2
b
2
)
.
Since
N
(
a
2
b
2
) is an
equivalence class and
a
1
b
1
∈
N
(
a
2
b
2
) then
N
(
a
1
b
1
) =
N
(
a
2
b
2
) (Theorem 9.2).
·
is associative
Let
a, b, c
∈
G.
Then
Na
(
NbNc
) =
Na
(
Nbc
) =
N
(
a
(
bc
)) =
N
((
ab
)
c
) =
N
(
ab
)
Nc
= (
NaNb
)(
Nc
)
1
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where we used the fact that multiplication in G is associative.
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 Winter '11
 PhanThuongCang
 Normal subgroup, Coset, G/N

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