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

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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