Prove that G - {0, 1, 2, 3, 4, 5, 6} is an abelian group of order 7 with respect to addition modulo 7.

Steps to Solve

1. Identify the set members We define the set ( G = { 7, 8, 9, 10, 11, 12, 13 } ), which is the result of adding 7 to each of the elements excluded from the original set ( {0, 1, 2, 3, 4, 5, 6} ) under addition modulo 7.

2. Verify closure property To check if the set is closed under addition modulo 7, we need to add every pair of elements in ( G ) and see if the result also belongs to ( G ).

For example:

• ( 7 + 7 = 14 \mod 7 = 0 ) (not in G)
• ( 7 + 8 = 15 \mod 7 = 1 ) (not in G)

Since the results of these sums are not found in ( G ), we need to adjust our understanding of the completeness of the modulo structure. Instead, consider the test cases solely on members:

• If ( a, b \in G ), then ( (a + b) \mod 7 ) will yield results in the range of 0 to 6, indicating that closure doesn't apply as expected without mapping back into the valid ( G ).

Instead, let's construct a proper cycling subset of ( { 1, 2, 3, 4, 5, 6 } ).

3. Check associativity For any ( a, b, c \in G ) if ( (a + b) + c = a + (b + c) ). This property holds for addition in integers and remains valid in modulo.

4. Identify the identity element The identity element under addition modulo 7 is 0. In our modified set ( G ), we can correlate ( 7 \equiv 0 \mod 7 ) which will confirm components related as an identity.

5. Prove existence of inverses For every ( a \in G ), there should exist ( b \in G ) such that ( a + b \equiv 0 \mod 7). Each number ( a = 7 + k ) for different k values will have its inverse because ( 6 + 1 \equiv 7 ).

6. Demonstrate commutativity For any ( a, b \in G ), ( a + b = b + a ). This property holds as addition of integers is inherently commutative.