For p 2 the situation is slightly more complicated

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For p = 2, the situation is slightly more complicated: Theorem 7.5 The group Z * 2 e is cyclic for e = 1 or 2 , but not for e 3 . For e 3 , Z * 2 e is isomorphic to the group Z 2 × Z 2 e - 2 . Before proving these two theorems, we need a few simple facts. Theorem 7.6 If p is prime and 0 < k < p , then the binomial coefficient ( p k ) is divisible by p . Proof. By definition p k = p ! k !( p - k )! . One sees that p divides the numerator, but for 0 < k < p , p does not divide the denominator. 2 Theorem 7.7 For e 1 , if a b (mod p e ) , then a p b p (mod p e +1 ) . Proof. We have a = b + cp e for some c Z . Thus, a p = b p + pb p - 1 cp e + dp 2 e for an integer d . It follows that a p b p (mod p e +1 ). 2 Theorem 7.8 Let e 1 and assume p e > 2 . If a 1 + p e (mod p e +1 ) , then a p 1 + p e +1 (mod p e +2 ) . Proof. By Theorem 7.7, a p (1 + p e ) p (mod p e +2 ). Expanding (1 + p e ) p , we have (1 + p e ) p = 1 + p · p e + p - 1 X k =2 p k p ek + p ep . Applying Theorem 7.6, all of the terms in the sum on k are divisible by p 1+2 e , and 1 + 2 e e + 2 for all e 1. For the term p ep , the assumption that p e > 2 means that either p 3 or e 2, which implies ep e + 2. 2 Now consider Theorem 7.4. Let p be odd and e > 1. Let x Z be chosen so that [ x mod p ] generates Z * p . Suppose the order of [ x mod p e ] Z * p e is m . Then as x m 1 (mod p e ) implies x m 1 (mod p ), it must be the case that p - 1 divides m , and so [ x m/ ( p - 1) mod p e ] has order exactly p - 1. By Theorem 4.29, if we find an integer y such that [ y mod p e ] has order p e - 1 , then [ x m/ ( p - 1) y mod p e ] has order ( p - 1) p e - 1 , and we are done. We claim that y = 1 + p does the job. Any integer between 0 and p e - 1 can be expressed as an e -digit number in base p ; for example, 49
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y = (0 · · · 0 1 1) p . If we compute successive p -th powers of y modulo p e , then by Theorem 7.8 we have: y rem p e = (0 · · · 0 1 1) p y p rem p e = ( * · · · * 1 0 1) p y p 2 rem p e = ( * · · · * 1 0 0 1) p . . . y p e - 2 rem p e = (1 0 · · · 0 1) p y p e - 1 rem p e = (0 · · · 0 1) p Here, “ * ” indicates an arbitrary digit. From this table of values, it is clear (c.f., Theorem 4.28) that [ y mod p e ] has order p e - 1 . That proves Theorem 7.4. Now consider Theorem 7.5. For e = 1 and e = 2, the theorem is clear. Suppose e 3. Consider the subgroup G Z * 2 e generated by [5 mod 2 e ]. Expressing integers between 0 and 2 e - 1 as e -digit binary numbers, and applying Theorem 7.8, we have: 5 rem 2 e = (0 · · · 0 1 0 1) 2 5 2 rem 2 e = ( * · · · * 1 0 0 1) 2 . . . 5 2 e - 3 rem 2 e = (1 0 · · · 0 1) 2 5 2 e - 2 rem 2 e = (0 · · · 0 1) 2 So it is clear (c.f., Theorem 4.28) that [5 mod 2 e ] has order 2 e - 2 . We claim that [ - 1 mod 2 e ] / G . If it were, then since it has order 2, and since any cyclic group of even order has precisely one element of order 2 (c.f., Theorem 4.24), it must be equal to [5 2 e - 3 mod 2 e ]; however, it is clear from the above calculation that 5 2 e - 3 6≡ - 1 (mod 2 e ). Let H Z * 2 e be the subgroup generated by [ - 1 mod 2 e ]. Then from the above, G H = { [1 mod 2 e ] } , and hence by Theorem 4.21, G × H is isomorphic to the subgroup G · H of Z * 2 e . But since the orders of G × H and Z * 2 e are equal, we must have G · H = Z * 2 e . That proves Theorem 7.5.
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