and is of order (p-1)p^{k-1} --> hence is a product G_k = A_k.B_k.
  with "core" subgroup |A_k|=p-1 (re: FST), and |B_k|= p^{k-1} = (p+1)*

A generator of G_2 will generate (if padded by msd zero's) also G_k
for every k>2. This is not always true for x* = G_1, where x may not
generate G_2.

Prime p=2 is the only prime for which G_k (k>1) is NOT cyclic,

but in fact a direct product: G_k = C2 x Cm,  where m= 2^{k-2}

 with 2-cycle C2={-1,1}

 and Cm={3*}: so 3 is a semi-primitive root of 1 mod 2^k (k>2).

see http://www.iae.nl/users/benschop/nfb0.dvi   (lem1.1)
   "Triplets and symmetries of Arithmetic mod p^k"
                   (incl. proof of FLTcase1 via FST_extension)

So strangely: the only interesting prime, from a practical point
of view (computer hardware: binary code mod 2^k) is also the only
one for which the primitive root problem is solved (and simple;-)

Ciao, Nico benschop -- http://www.iae.nl/users/benschop