Prove that the set G ={1, 2, 3, 4, 5, 6} is an abelian group under multiplication modulo 7.

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written 7.9 years ago by

teamques10 ★ 68k

• modified 7.9 years ago

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

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$(G_1)$ 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.

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