Question

Let $\mathit{G}=\left\{\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{,}\mathrm{2}\mathrm{,}\mathrm{3}\mathrm{,}\mathrm{4}\mathrm{,}\mathrm{5}\mathrm{,}\mathrm{6}\mathrm{,}\mathrm{7}\right\}$ and assume G is a group under an operation * with these properties:

(i) $\mathit{a}\mathbf{*}\mathit{b}\mathbf{\le }\mathit{a}\mathbf{+}\mathit{b}$for all $\mathit{a}\mathbf{,}\mathit{b}\in \mathit{G}$;

(ii) $\mathit{a}\mathbf{*}\mathit{a}=0$for all $\mathit{a}\in \mathit{G}$.

Write out the operation table for G. [Hint: Exercises 27 and 28 may help.]

Short answer

The operation table for G is as follows.

*	0	1	2	3	4	5	6	7
0	0	1	2	3	4	5	6	7
1	1	0	3	2	5	4	7	6
2	2	3	0	1	6	7	5	4
3	3	2	1	0	7	6	4	5
4	4	5	6	7	0	3	1	2
5	5	4	7	6	3	0	2	1
6	6	7	5	4	1	2	0	3
7	7	6	4	5	2	1	3	0

Step-by-step solution

Given information

Consider a set $G=\left\{0,1,2,3,4,5,6,7\right\}$. According to the problem, G is a group under the operation * with the following properties:

First; $a*b\le a+b$for all $a,b\in G$

Second; $a*a=0$for all$a\in G$

Write out the operation table for G

The given condition force $0$to be the identity element. So start the table as follows.

*	0	1	2	3	4	5	6	7
0	0	1	2	3	4	5	6	7
1	1	0
2	2		0
3	3			0
4	4				0
5	5					0
6	6						0
7	7							0

From the condition (i) and exercise 28, 1 * 2 = 3 and 2 * 1 = 3. So, fill the table as follows.

*	0	1	2	3	4	5	6	7
0	0	1	2	3	4	5	6	7
1	1	0	3
2	2	3	0
3	3			0
4	4				0
5	5					0
6	6						0
7	7							0

From the condition (i) and exercise 28, 3 * 1 is either 2 or 4 , but associative property gives that if 3 * 1 = 2 and 1 * 3 = 2 . So, fill the table as follows.

*	0	1	2	3	4	5	6	7
0	0	1	2	3	4	5	6	7
1	1	0	3	2
2	2	3	0
3	3	2		0
4	4				0
5	5					0
6	6						0
7	7							0

Similarly, fill the table as follows.

*	0	1	2	3	4	5	6	7
0	0	1	2	3	4	5	6	7
1	1	0	3	2	5	4	7	6
2	2	3	0	1	6	7	5	4
3	3	2	1	0	7	6	4	5
4	4	5	6	7	0	3	1	2
5	5	4	7	6	3	0	2	1
6	6	7	5	4	1	2	0	3
7	7	6	4	5	2	1	3	0