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structure of (Z/nZ)× as an abelian group∗ rm50† 2013-03-22 2:23:17 The automorphism group of the cyclic group Cn ∼ = Z/nZ is (Z/nZ)× . This × article determines the structure of (Z/nZ) as an abelian group. Theorem 1. Let n ≥ 2 be an integer whose factorization is n = pa1 1 pa2 2 . . . par r where the pi are distinct primes. Then: 1. (Z/nZ)× ∼ = (Z/pa1 1 Z)× × (Z/pa2 2 Z)× × · · · × (Z/par r Z)× 2. (Z/pk Z)× is a cyclic group of order pk−1 (p − 1) for all odd primes p. 3. (Z/2k Z)× is the direct product of a cyclic group of order 2 and a cyclic group of order 2k−2 for k ≥ 2. Corollary 2. Aut(Cn ) ∼ = (Z/nZ)× is cyclic if and only if n = 2, 4, pk , or 2pk for p an odd prime and k ≥ 0 an integer. Proof. (of theorem) (1): This is a restatement of the Chinese Remainder Theorem. (2): Note first that the result is clear for k = 1, since then (Z/pZ)× is the multiplicative group of the finite field Z/pZ and thus is cyclic (any finite subgroup of the multiplicative group of a field is cyclic). Also, (Z/pk Z)× = φ(pk ) = pk−1 (p − 1). Since (Z/pk Z)× is abelian, it is the direct product of its q-primary components for each prime q | φ(pk ); we will show that each of those q-primary components is cyclic. For q = p, it suffices to find an element of (Z/pk Z)× of order pk−1 . 1 + p is such an element; see Lemma ?? below. For q 6= p, consider the map Z/pk Z → Z/pZ : a + (pk ) 7→ a + (p) i.e. the reduction-by-p map. This is a ring homomorphism; restricting it to (Z/pk Z)× gives a surjective group homomorphism π : (Z/pk Z)× → (Z/pZ)× . Since |(Z/pZ)× | = p − 1, it follows that the kernel of π has order pk−1 . Thus ∗ hStructureOfmathbbZnmathbbZtimesAsAnAbelianGroupi created: h2013-03-2i by: hrm50i version: h41477i Privacy setting: h1i hTheoremi h20A05i h20E36i h20E34i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 for q 6= p, the q-primary component of (Z/pk Z)× must map isomorphically into (Z/pZ)× by order considerations. But (Z/pZ)× is cyclic, so the q-primary component is as well. Thus each q-primary component of (Z/pk Z)× is cyclic and thus (Z/pk Z)× is also cyclic. (3): The result is true for k = 2, when (Z/22 Z)× ∼ = V4 , the Klein 4-group. So assume k ≥ 3. 5 has exact order 2k−2 in (Z/2k Z)× (see Lemma ?? below). Also k−3 k−3 by that lemma, 52 6= −1 is (Z/2k Z)× , so that 52 and −1 are two distinct elements of order 2. Thus (Z/2k Z)× is not cyclic, but has a cyclic subgroup of order 2k−2 ; the result follows. Proof. (of Corollary) (⇐) is clear, since (Z/C2 Z)× ∼ = {1} (Z/Cpk Z)× ∼ = Cpk−1 (p−1) (Z/C4 Z)× ∼ = C2 (Z/C2pk Z)× ∼ = (Z/C2 Z)× × (Z/Cpk Z)× ∼ = Cpk−1 (p−1) (⇒): Assume (Z/nZ)× is cyclic. If n is a power of 2, then by the theorem, it must be either 2 or 4. Otherwise, if n has two distinct odd prime factors p, q, then (Z/nZ)× contains the direct product (Z/pr Z)× × (Z/q s Z)× . But the orders of these two factor groups are both even (they are φ(pr ) = pr−1 (p−1) and φ(q s ) = q s−1 (q − 1) respectively), so their direct product is not cyclic. Thus n can have at most one odd prime as a factor, so that n = 2m pk for some integers m, k, and (Z/Cn Z)× = (Z/C2m Z)× × (Z/Cpk Z)× But the order of (Z/Cpk Z)× is even, so that (since the order of (Z/C2m Z)× is also even for m ≥ 2) we must have m = 0 or 1, so that n = pk or 2pk . The above proof used the following lemmas, which we now prove: Lemma 3. Let p be an odd prime and k > 0 a positive integer. Then 1 + p has exact order pk−1 in the multiplicative group (Z/pk Z)× . Proof. The result is obvious for k = 1, so we assume k ≥ 2. By the binomial theorem, pn n X p pn (1 + p) = 1 + pi i i=1 Write ordp (m) for the largest power of a prime p dividing m. Then by a theorem on divisibility of prime-power binomial coefficients, n p ordp pi = n + i − ordp (i) i Now, i − ordp (i) is 1 if i = 1, and is at least 2 for i > 1 (since p ≥ 3). We thus get n (1 + p)p = 1 + pn+1 + rpn+2 , r ∈ Z 2 Setting n = k −1 gives (1+p)p 1 + pk−1 6≡ 1 (pk ). k−1 ≡ 1 (pk ); setting n = k −2 gives (1+p)p k−2 ≡ Lemma 4. For k ≥ 3, 5 has exact order 2k−2 in the multiplicative group k−3 (Z/2k Z)× (which has order 2k−1 ). Additionally, 52 6≡ −1 (2k ). Proof. The proof of this lemma is essentially identical to the proof of the preceding lemma. Again by the binomial theorem, n 2n 2 2n 5 = (1 + 2 ) = 2 n X 2 i=1 Then ord2 i 22i 2n i 2 = n + 2i − ord2 (i) i Now, 2i − ord2 (i) is 2 if i = 1, and is at least 3 for i > 1. We thus get n 52 = 1 + 2n+2 + r2n+3 , k−2 r∈Z k−3 Setting n = k−2 gives 52 ≡ 1 (2k ); setting n = k−3 gives 52 k k−1 ±1 (2 ). (Note that 1 + 2 6≡ −1 since k ≥ 3). ≡ 1+2k−1 6≡ References [1] Dummit, D., Foote, R.M., Abstract Algebra, Third Edition, Wiley, 2004. 3

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