Compute 3 3 5 3 and 5 31 in U7 Z4Solutionhere U7 is multipl

Compute (3, 3) · (5, 3) and (5, 3)^-1 in U(7) Z₄

Solution

here, U(7) is multiplicative group consisting {0,1,2,3,4,5,6} operation is Xn denotes the remainder obtained after multiplication with respect to 7.

Z4 is additive group consisting {0,1,2,3}

hence (3,3).(5,3)=(3.5,3.3)=(1,2)

the other question is not clearly stated