**1 Answer**

written 7.4 years ago by |

Since set is finite, we prepare the following multiplication table to examine the group axioms.

$(G_1)$ All the entries in the table are elements of G. Therefore G is closed with respect to multiplication modulo 7.

$(G_2)$ Multiplication modulo 7 is associative.

$(G_3)$ Since first row of the is identical to the row of elements of G in the horizontal border, the element to the left of first row in vertical border is identity element i.e., 1 is identity element in G with respect to multiplication mod 7.

$(G_4)$ From the table it is obvious that inverses of 1,2,3,4,5,6 are 1,4,5,2,3 and 6 respectively. Hence inverse of each element in G exists.

$(G_5)$ The composition is commutative because the elements equidistant from principal diagonal are equal each to each.

The set G has six element. Hence, $(G,x_7)$ is a finite abelian group of order 6.