3.1. DIRECT PRODUCTS
149
Proposition 3.1.5.
(a)
Suppose
M
and
N
are normal subgroups of
G
, and
M
\
N
D
f
e
g
. Then for all
m
2
M
and
n
2
N
,
mn
D
nm
.
(b)
MN
D f
mn
W
m
2
M
and
n
2
N
g
is a subgroup and
.m;n/
7!
mn
is an isomorphism of
M
±
N
onto
MN
.
(c)
If
MN
D
G
, then
G
Š
M
±
N
.
Proof.
We have
mn
D
nm
,
mnm
±
1
n
±
1
D
e
. Observe that
mnm
±
1
n
±
1
D
.mnm
±
1
/n
2
N;
since
N
is normal. Likewise,
mnm
±
1
n
±
1
D
m.nm
±
1
n
±
1
/
2
M;
since
M
is normal. Since the intersection of
M
and
N
is trivial, it follows
that
mnm
±
1
n
±
1
D
e
.
It is now easy to check that
MN
is a subgroup, and that
.m;n/
7!
mn
is a homomorphism of
M
±
N
onto
MN
. The homomorphism is injective,
because if
mn
D
e
, then
m
D
n
±
1
2
M
\
N
D f
e
g
.
Assertion (c) is evident.
n
Notation 3.1.6.
When
G
has subgroups
N
and
M
such that
N
\
M
D f
e
g
and
NM
D
G
, we will write
G
D
N
±
M
to convey
that
G
Š
N
±
M
and
N
and
M
are subgroups of
G
. See Remark
3.1.9
.
Example 3.1.7.
Let’s look at Example
3.1.3
again “from the inside” of
Z
ab
. The subgroup
h
ŒaŁ
i
of
Z
ab
generated by
ŒaŁ
has order
b
and the
subgroup
h
ŒbŁ
i
has order
a
. These two subgroups have trivial intersec
tion because the order of the intersection must divide the orders of both
subgroups, and the orders of the subgroups are relatively prime. The two
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 Fall '08
 EVERAGE
 Algebra, subgroups, Za Zb

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