Q: in a group{1,2,.....16} under the operation of multiplication modulo 17,the order of element 3 is???
just multiply
$\displaystyle 3=3 (mod 17)$
$\displaystyle 3^2=9 (mod 17)$
$\displaystyle 3^3=9*3=27 \equiv 10 (mod 17)$
$\displaystyle 3^4=10*3=30 \equiv 13 (mod 17)$
$\displaystyle 3^5=13*3=39 \equiv 5 (mod 17)$
$\displaystyle 3^6=5*3=15 \equiv -2 (mod 17)$
$\displaystyle 3^7= -2*3 = -6 (mod 17)$
$\displaystyle 3^8=-6*3 = -18 \equiv -1 (mod 17)$
$\displaystyle 3^{16}=(3^8)^2=(-1)^2=1 (mod 17)$
order is 16 if I did it all right, you can do this on your own, this says 3 is a generator for $\displaystyle \left(\frac{\mathbb{Z}}{(17\mathbb{Z})}\right)^{\t imes}$