For a finite abelian group one can completely specify the gr

For a finite abelian group, one can completely specify the group by writing down the group operation table. For instance, Example 2.7 presented an addition table for Z_6. Write down group operation tables for the following finite abelian groups: Z_5, Z*_5, and Z_3 times Z*_4. Show that the group operation table for every finite abelian group is a Latin square; that is, each element of the group appears exactly once in each row and column. Below is an addition table for an abelian group that consists of the elements {a, b, c, d}; however, some entries are missing. Fill in the missing entries.

Solution

(a) Group operation table for Z₅

+

0

1

2

3

4

0

0

1

2

3

4

1

1

2

3

4

0

2

2

3

4

0

1

3

3

4

0

1

2

4

4

0

1

2

3

Group operation table for Z₅^*

.

1

2

3

4

1

1

2

3

4

2

2

4

1

3

3

3

1

4

2

4

4

3

2

1

Z₃ x Z₄^* = {(a, b) : a Z₃, b Z₄^*}. The binary operation o is defined as (a, b)o(c, d) = (a+c, b.d) where + is the binary operation (addition modulo 3) for the group Z₃ and . is the binary operation (multiplication modulo 4) for the group Z₄^*

Group operation table for Z₃ x Z₄^*

o

(0, 1)

(0, 3)

(1, 1)

(1, 3)

(2, 1)

(2, 3)

(0, 1)

(0, 1)

(0, 3)

(1, 1)

(1, 3)

(2, 1)

(2, 3)

(0, 3)

(0, 3)

(0, 1)

(1, 3)

(1, 1)

(2, 3)

(2, 1)

(1, 1)

(1, 1)

(1, 3)

(2, 1)

(2, 3)

(0, 1)

(0, 3)

(1, 3)

(1, 3)

(1, 1)

(2, 3)

(2, 1)

(0, 3)

(0, 1)

(2, 1)

(2, 1)

(2, 3)

(0, 1)

(0, 3)

(1, 1)

(1, 3)

(2, 3)

(2, 3)

(2, 1)

(0, 3)

(0, 1)

(1, 3)

(1, 1)

(c) G is an abelian group, G = {a, b, c, d}. The group operation table (incomplete) is defined as

+

a

b

c

d

a

a

b

b

a

c

a

d

a + a = a, a + b = b + a = b, b + b = a, c + c = a

a + b = b + a = b implies that a is the identity element of G.

b + b = a, c + c = a implies that both b and c are of order 2.

As the order of G is 4, and it has two elements b and c of order 2, then d is also of order 2. So, d + d = a. With this information we can partially fill the table.

+

a

b

c

d

a

a

b

c

d

b

b

a

c

c

a

d

d

a

Now, b + c a (as a already appeared in the row)

b + c b, c (as both b and c are non-identity element)

So, b + c = c + b = d

Similarly we get, b + d = d + b = c

And, c + d = d + c = b

Hence, the complete group operation table is

+

a

b

c

d

a

a

b

c

d

b

b

a

d

c

c

c

d

a

b

d

d

c

b

a

(b) For any finite abelian group G, the group operation table (also known as Cayley table) is a Latin square, i.e. each element of the group appears exactly once in each row and column.

Proof: If the group G has n elements, then its Cayley table is, by definition, an n × n array, in which the entries are labelled by the n elements of G. It remains to show that each element g G appears exactly once in each row and in each column. We will show that g appears exactly once in each row. The argument for columns is similar.

Suppose first that g appears twice in row x.

y

z

x

g

g

Then there are two distinct elements y, z G such that x y = g = x z. Let x^-1 be the inverse of x in (G, ). Then

y = e_G y = (x^-1 x) y = x^-1 (x y) = x^-1 g = x^-1 (x z)=(x^-1 x) z = e_G z = z, contrary to the assumption that y, z are distinct.

Hence g cannot appear twice in any row of the Cayley table. A similar argument applies to any other element of the group, so no element appears twice in the same row. But there are n entries in each row, and n possible labels for the entries. By the pigeonhole principle, if some label did not occur in a given row, then some other label would have to occur twice, which we have seen is impossible. Hence each element of G occurs exactly once in each row of the table. (Proved)

+	0	1	2	3	4
0	0	1	2	3	4
1	1	2	3	4	0
2	2	3	4	0	1
3	3	4	0	1	2
4	4	0	1	2	3