We can define a special group product, x, between two cyclic groups, Z, and Zm, on the set Zn X Zm, where the group operation, , is given by (i, x) e (j,y) = (it(-1)"j, xty), i,jeZn, X,y EZm, where arithmetic in the first coordinate is (mod n) and (mod m) in the second. So, for example, in Z1 x Z2: (2, 1) @ (3, 0) = (2 + (-1) 3, 1 + 0) = (2 - 3, 1) = (-1, 1) = (3, 1) (2, 1) (3, 1) = (2 + (-1) 3, 1 + 1) = (2 - 3, 0) = (-1,0) = (3, 0) Recall that Dn = {e, r, ..., po-1, ra, ..., r-la}, where r| = n, Jal = 2, and ra = ar-. Show that Zn X Z2 ~ Dn