1Groups and homomorphisms

IA Groups



1.5 Direct products of groups
Recall that if we have to sets
X, Y
, then we can obtain the product
X × Y
=
{(x, y) : x X, y Y }. We can do the same if X and Y are groups.
Definition
(Direct product of groups)
.
Given two groups (
G,
) and (
H,
),
we can define a set
G × H
=
{
(
g, h
) :
g G, h H}
and an operation
(a
1
, a
2
) (b
1
, b
2
) = (a
1
b
1
, a
2
b
2
). This forms a group.
Why would we want to take the product of two groups? Suppose we have
two independent triangles. Then the symmetries of this system include, say
rotating the first triangle, rotating the second, or rotating both. The symmetry
group of this combined system would then be D
6
× D
6
.
Example.
C
2
× C
2
= {(0, 0), (0, 1), (1, 0), (1, 1)}
= {e, x, y, xy} with everything order 2
= hx, y | x
2
= y
2
= e, xy = yxi
Proposition. C
n
× C
m
=
C
nm
iff hcf(m, n) = 1.
Proof.
Suppose that
hcf
(
m, n
) = 1. Let
C
n
=
hai
and
C
m
=
hbi
. Let
k
be the
order of (
a, b
). Then (
a, b
)
k
= (
a
k
, b
k
) =
e
. This is possible only if
n | k
and
m | k
, i.e.
k
is a common multiple
n
and
m
. Since the order is the minimum
value of k that satisfies the above equation, k = lcm(n, m) =
nm
hcf(n,m)
= nm.
Now consider
h
(
a, b
)
i C
n
× C
m
. Since (
a, b
) has order
nm
,
h
(
a, b
)
i
has
nm
elements. Since
C
n
× C
m
also has
nm
elements,
h
(
a, b
)
i
must be the whole of
C
n
× C
m
. And we know that h(a, b)i
=
C
nm
. So C
n
× C
m
=
C
nm
.
On the other hand, suppose
hcf
(
m, n
)
6
= 1. Then
k
=
lcm
(
m, n
)
6
=
mn
. Then
for any (
a, b
)
C
n
× C
m
,we have (
a, b
)
k
= (
a
k
, b
k
) =
e
. So the order of any (
a, b
)
is at most
k < mn
. So there is no element of order
mn
. So
C
n
× C
m
is not a
cyclic group of order nm.
Given a complicated group
G
, it is sometimes helpful to write it as a product
H × K
, which could make things a bit simpler. We can do so by the following
theorem:
Proposition
(Direct product theorem)
.
Let
H
1
, H
2
G
. Suppose the following
are true:
(i) H
1
H
2
= {e}.
(ii) (a
i
H
i
) a
1
a
2
= a
2
a
1
.
(iii) (a G)(a
i
H
i
) a = a
1
a
2
. We also write this as G = H
1
H
2
.
Then G
=
H
1
× H
2
.
Proof.
Define
f
:
H
1
×H
2
G
by
f
(
a
1
, a
2
) =
a
1
a
2
. Then it is a homomorphism
since
f((a
1
, a
2
) (b
1
, b
2
)) = f(a
1
b
1
, a
2
b
2
)
= a
1
b
1
a
2
b
2
= a
1
a
2
b
1
b
2
= f(a
1
, a
2
)f(b
1
, b
2
).
Surjectivity follows from (iii). We’ll show injectivity by showing that the kernel
is
{e}
. If
f
(
a
1
, a
2
) =
e
, then we know that
a
1
a
2
=
e
. Then
a
1
=
a
1
2
. Since
a
1
H
1
and
a
1
2
H
2
, we have
a
1
=
a
1
2
H
1
H
2
=
{e}
. Thus
a
1
=
a
2
=
e
and ker f = {e}.