structure of (/n)× as an abelian group

The automorphism groupMathworldPlanetmath of the cyclic groupMathworldPlanetmath Cn/n is (/n)×. This article determines the structureMathworldPlanetmath of (/n)× as an abelian groupMathworldPlanetmath.

Theorem 1.

Let n2 be an integer whose factorization is n=p1a1p2a2prar where the pi are distinct primes. Then:

  1. 1.


  2. 2.

    (/pk)× is a cyclic group of order pk-1(p-1) for all odd primes p.

  3. 3.

    (/2k)× is the direct productMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of a cyclic group of order 2 and a cyclic group of order 2k-2 for k2.

Corollary 2.

Aut(Cn)(/n)× is cyclic if and only if n=2,4,pk, or 2pk for p an odd prime and k0 an integer.


(of theorem)
(1): This is a restatement of the Chinese Remainder TheoremMathworldPlanetmathPlanetmathPlanetmath.

(2): Note first that the result is clear for k=1, since then (/p)× is the multiplicative groupMathworldPlanetmath of the finite field /p and thus is cyclic (any finite subgroup of the multiplicative group of a field is cyclic). Also, |(/pk)×|=ϕ(pk)=pk-1(p-1). Since (/pk)× is abelian, it is the direct product of its q-primary componentsPlanetmathPlanetmath for each prime qϕ(pk); we will show that each of those q-primary components is cyclic. For q=p, it suffices to find an element of (/pk)× of order pk-1. 1+p is such an element; see Lemma 3 below. For qp, consider the map


i.e. the reduction-by-p map. This is a ring homomorphismMathworldPlanetmath; restricting it to (/pk)× gives a surjectivePlanetmathPlanetmath group homomorphismMathworldPlanetmath π:(/pk)×(/p)×. Since |(/p)×|=p-1, it follows that the kernel of π has order pk-1. Thus for qp, the q-primary component of (/pk)× must map isomorphically into (/p)× by order considerations. But (/p)× is cyclic, so the q-primary component is as well.

Thus each q-primary component of (/pk)× is cyclic and thus (/pk)× is also cyclic.

(3): The result is true for k=2, when (/22)×V4, the Klein 4-group. So assume k3. 5 has exact order 2k-2 in (/2k)× (see Lemma 4 below). Also by that lemma, 52k-3-1 is (/2k)×, so that 52k-3 and -1 are two distinct elements of order 2. Thus (/2k)× is not cyclic, but has a cyclic subgroup of order 2k-2; the result follows. ∎


(of Corollary)
() is clear, since


(): Assume (/n)× 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 (/n)× contains the direct product (/pr)××(/qs)×. But the orders of these two factor groups are both even (they are ϕ(pr)=pr-1(p-1) and ϕ(qs)=qs-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=2mpk for some integers m,k, and


But the order of (/Cpk)× is even, so that (since the order of (/C2m)× is also even for m2) 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/pkZ)×.


The result is obvious for k=1, so we assume k2. By the binomial theoremMathworldPlanetmath,


Write ordp(m) for the largest power of a prime p dividing m. Then by a theorem on divisibility of prime-power binomial coefficients,


Now, i-ordp(i) is 1 if i=1, and is at least 2 for i>1 (since p3). We thus get


Setting n=k-1 gives (1+p)pk-11(pk); setting n=k-2 gives (1+p)pk-21+pk-11(pk). ∎

Lemma 4.

For k3, 5 has exact order 2k-2 in the multiplicative group (Z/2kZ)× (which has order 2k-1). Additionally, 52k-3-1(2k).


The proof of this lemma is essentially identical to the proof of the preceding lemma. Again by the binomial theorem,




Now, 2i-ord2(i) is 2 if i=1, and is at least 3 for i>1. We thus get


Setting n=k-2 gives 52k-21(2k); setting n=k-3 gives 52k-31+2k-1±1(2k). (Note that 1+2k-1-1 since k3). ∎


Title structure of (/n)× as an abelian group
Canonical name StructureOfmathbbZnmathbbZtimesAsAnAbelianGroup
Date of creation 2013-03-22 18:42:40
Last modified on 2013-03-22 18:42:40
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 4
Author rm50 (10146)
Entry type Theorem
Classification msc 20A05
Classification msc 20E36
Classification msc 20E34