Prove that the set G = {0, 1, 2, 3, 4, 5, 6} is a finite abelian group of order 7 with respect to multiplication modulo 7.

To prove that the set G = {0, 1, 2, 3, 4, 5, 6} is a finite abelian group of order 7 with respect to multiplication modulo 7, we need to show that it satisfies the four group axioms: closure, associativity, identity, and invertibility. Also, since it's an abelian group, it should satisfy the commutative property.

1. Closure: The operation is closed if the product of any two elements in the set is also in the set. In this case, the operation is multiplication modulo 7. For any two elements a and b in G, (a*b) mod 7 is also in G. For example, (3*5) mod 7 = 15 mod 7 = 1, which is in G.

2. Associativity: The operation is associative if for any three elements a, b, and c in G, (a*b)c = a(b*c). In this case, for any a, b, and c in G, ((a*b) mod 7)c mod 7 = a((b*c) mod 7) mod 7. For example, ((2*3) mod 7)4 mod 7 = 2((3*4) mod 7) mod 7 = 6, which proves associativity.

3. Identity: There is an identity element in G, which when operated with any element in G, leaves the element unchanged. In this case, the identity element is 1 because for any a in G, (a*1) mod 7 = a. For example, (3*1) mod 7 = 3.

4. Invertibility: Every element in G has an inverse in G. An element a in G has an inverse b in G if (a*b) mod 7 = 1. For example, the inverse of 3 is 5 because (3*5) mod 7 = 1.

5. Commutativity: The operation is commutative if for any two elements a and b in G, a*b = b*a. In this case, for any a and b in G, (a*b) mod 7 = (b*a) mod 7. For example, (2*3) mod 7 = (3*2) mod 7 = 6.

Here is a table showing the multiplication modulo 7 for all elements in G:

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

From the table, we can see that all the group axioms and the commutative property are satisfied. Therefore, G = {0, 1, 2, 3, 4, 5, 6} is a finite abelian group of order 7 with respect to multiplication modulo 7.