Math 302
Modern Algebra
Spring 2009
Homework Solutions
3.3
ISOMORPHISMS AND HOMOMORPHISMS
1.
Write out the addition and multiplication tables for
Z
6
and for
Z
2
×
Z
3
.
Use them to show that
Z
6
∼
=
Z
2
×
Z
3
.
The addition and multiplication tables for
Z
6
are
+
0
1
2
3
4
5
0
0
1
2
3
4
5
1
1
2
3
4
5
0
2
2
3
4
5
0
1
3
3
4
5
0
1
2
4
4
5
0
1
2
3
5
5
0
1
2
3
4
·
0
1
2
3
4
5
0
0
0
0
0
0
0
1
0
1
2
3
4
5
2
0
2
4
0
2
4
3
0
3
0
3
0
3
4
0
4
2
0
4
2
5
0
5
4
3
2
1
The addition and multiplication tables for
Z
2
×
Z
3
are
+
(0
,
0)
(1
,
1)
(0
,
2)
(1
,
0)
(0
,
1)
(1
,
2)
(0
,
0)
(0
,
0)
(1
,
1)
(0
,
2)
(1
,
0)
(0
,
1)
(1
,
2)
(1
,
1)
(1
,
1)
(0
,
2)
(1
,
0)
(0
,
1)
(1
,
2)
(0
,
0)
(0
,
2)
(0
,
2)
(1
,
0)
(0
,
1)
(1
,
2)
(0
,
0)
(1
,
1)
(1
,
0)
(1
,
0)
(0
,
2)
(1
,
2)
(0
,
0)
(1
,
1)
(0
,
2)
(0
,
1)
(0
,
1)
(1
,
2)
(0
,
0)
(1
,
1)
(0
,
2)
(1
,
0)
(1
,
2)
(1
,
2)
(0
,
0)
(1
,
1)
(0
,
2)
(1
,
0)
(0
,
1)
·
(0
,
0)
(1
,
1)
(0
,
2)
(1
,
0)
(0
,
1)
(1
,
2)
(0
,
0)
(0
,
0)
(0
,
0)
(0
,
0)
(0
,
0)
(0
,
0)
(0
,
0)
(1
,
1)
(0
,
0)
(1
,
1)
(0
,
2)
(1
,
0)
(0
,
1)
(0
,
2)
(0
,
2)
(0
,
0)
(0
,
2)
(0
,
1)
(0
,
0)
(0
,
2)
(0
,
1)
(1
,
0)
(0
,
0)
(1
,
0)
(0
,
0)
(1
,
0)
(0
,
0)
(1
,
0)
(0
,
1)
(0
,
0)
(0
,
1)
(0
,
2)
(0
,
0)
(0
,
1)
(0
,
2)
(1
,
2)
(0
,
0)
(1
,
2)
(0
,
1)
(1
,
0)
(0
,
2)
(1
,
1)
In order to determine an isomorphism
f
:
Z
6
→
Z
2
×
Z
3
, first note that
f
must map the additive
and multiplicative identities of
Z
6
to the additive and multiplicative identities of
Z
2
×
Z
3
, so
f
(0) = (0
,
0)
and
f
(1) = (1
,
1)
.
Since
f
must preserve addition, then we must have
f
(2) =
f
(1 + 1) =
f
(1) +
f
(1) = (1
,
1) + (1
,
1) = (0
,
2)
f
(3) =
f
(1 + 2) =
f
(1) +
f
(2) = (1
,
1) + (0
,
2) = (1
,
0)
f
(4) =
f
(1 + 3) =
f
(1) +
f
(3) = (1
,
1) + (1
,
0) = (0
,
1)
f
(5) =
f
(1 + 4) =
f
(1) +
f
(4) = (1
,
1) + (0
,
1) = (1
,
2)
This defines our function. To verify that it is in fact an isomorphism, we can compare the addition
and multiplication tables with the correspondence
0
↔
(0
,
0)
,
1
↔
(1
,
1)
,
2
↔
(0
,
2)
,
3
↔
(1
,
0)
,
4
↔
(0
,
1)
,
5
↔
(1
,
2)
.
If we replace the elements in the tables for
Z
6
with the corresponding elements in
Z
2
×
Z
3
, we should
get the tables for
Z
2
×
Z
3
shown. This shows that this function is an isomorphism.
33