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
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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.
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