MIke Reid writes:

<<
when we work modulo  2^n  (n > 2), this means that  (Z / 2^n Z)^*  is isomorphic to a product of two cyclic groups, C_2 x C_(2^n-2)  with generators  -1  and  5
>>

By coincidence, just today I was reviewing the proof that for an odd prime, (Z/p^nZ)* is cyclic, and it's amazing  how many websites "prove" it without noting the exceptions for p = 2, n > 2.

--Dan

--Dan