Multiplicative group of integers modulo n
Multiplicative group of integers modulo n
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Multiplicative group of integers modulo n

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In modular arithmetic, the integers coprime (relatively prime) to n from the set of n non-negative integers form a group under multiplication modulo n, called the multiplicative group of integers modulo n. Equivalently, the elements of this group can be thought of as the congruence classes, also known as residues modulo n, that are coprime to n. Hence another name is the group of primitive residue classes modulo n. In the theory of rings, a branch of abstract algebra, it is described as the group of units of the ring of integers modulo n. Here units refers to elements with a multiplicative inverse, which, in this ring, are exactly those coprime to n.

This group, usually denoted , is fundamental in number theory. It is used in cryptography, integer factorization, and primality testing. It is an abelian, finite group whose order is given by Euler's totient function: For prime n the group is cyclic, and in general the structure is easy to describe, but no simple general formula for finding generators is known.

Group axioms

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It is a straightforward exercise to show that, under multiplication, the set of congruence classes modulo n that are coprime to n satisfy the axioms for an abelian group.

Indeed, a is coprime to n if and only if gcd(a, n) = 1. Integers in the same congruence class ab (mod n) satisfy gcd(a, n) = gcd(b, n); hence one is coprime to n if and only if the other is. Thus the notion of congruence classes modulo n that are coprime to n is well-defined.

Since gcd(a, n) = 1 and gcd(b, n) = 1 implies gcd(ab, n) = 1, the set of classes coprime to n is closed under multiplication.

Integer multiplication respects the congruence classes; that is, aa' and bb' (mod n) implies aba'b' (mod n). This implies that the multiplication is associative, commutative, and that the class of 1 is the unique multiplicative identity.

Finally, given a, the multiplicative inverse of a modulo n is an integer x satisfying ax ≡ 1 (mod n). It exists precisely when a is coprime to n, because in that case gcd(a, n) = 1 and by Bézout's lemma there are integers x and y satisfying ax + ny = 1. Notice that the equation ax + ny = 1 implies that x is coprime to n, so the multiplicative inverse belongs to the group.

Notation

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The set of (congruence classes of) integers modulo n with the operations of addition and multiplication is a ring. It is denoted   or    (the notation refers to taking the quotient of integers modulo the ideal or consisting of the multiples of n). Outside of number theory the simpler notation is often used, though it can be confused with the p-adic integers when n is a prime number.

The multiplicative group of integers modulo n, which is the group of units in this ring, may be written as (depending on the author)         (for German Einheit, which translates as unit), , or similar notations. This article uses

The notation refers to the cyclic group of order n. It is isomorphic to the group of integers modulo n under addition. Note that or may also refer to the group under addition. For example, the multiplicative group for a prime p is cyclic and hence isomorphic to the additive group , but the isomorphism is not obvious.

Structure

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The order of the multiplicative group of integers modulo n is the number of integers in coprime to n. It is given by Euler's totient function: (sequence A000010 in the OEIS). For prime p, .

Cyclic case

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The group is cyclic if and only if n is 1, 2, 4, pk or 2pk, where p is an odd prime and k > 0. For all other values of n the group is not cyclic.[1][2][3] This was first proved by Gauss.[4]

This means that for these n:

where

By definition, the group is cyclic if and only if it has a generator g; that is, the powers give all possible residues modulo n coprime to n (the first powers give each exactly once). A generator of is called a primitive root modulo n.[5] If there is any generator, then there are of them.

Powers of 2

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Modulo 1 any two integers are congruent, i.e., there is only one congruence class, [0], coprime to 1. Therefore, is the trivial group with φ(1) = 1 element. Because of its trivial nature, the case of congruences modulo 1 is generally ignored and some authors choose not to include the case of n = 1 in theorem statements.

Modulo 2 there is only one coprime congruence class, [1], so is the trivial group.

Modulo 4 there are two coprime congruence classes, [1] and [3], so the cyclic group with two elements.

Modulo 8 there are four coprime congruence classes, [1], [3], [5] and [7]. The square of each of these is 1, so the Klein four-group.

Modulo 16 there are eight coprime congruence classes [1], [3], [5], [7], [9], [11], [13] and [15]. is the 2-torsion subgroup (i.e., the square of each element is 1), so is not cyclic. The powers of 3, are a subgroup of order 4, as are the powers of 5,   Thus

The pattern shown by 8 and 16 holds[6] for higher powers 2k, k > 2: is the 2-torsion subgroup, so cannot be cyclic, and the powers of 3 are a cyclic subgroup of order 2k − 2, so:

General composite numbers

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By the fundamental theorem of finite abelian groups, the group is isomorphic to a direct product of cyclic groups of prime power orders.

More specifically, the Chinese remainder theorem[7] says that if then the ring is the direct product of the rings corresponding to each of its prime power factors:

Similarly, the group of units is the direct product of the groups corresponding to each of the prime power factors:

For each odd prime power the corresponding factor is the cyclic group of order , which may further factor into cyclic groups of prime-power orders. For powers of 2 the factor is not cyclic unless k = 0, 1, 2, but factors into cyclic groups as described above.

The order of the group is the product of the orders of the cyclic groups in the direct product. The exponent of the group; that is, the least common multiple of the orders in the cyclic groups, is given by the Carmichael function (sequence A002322 in the OEIS). In other words, is the smallest number such that for each a coprime to n, holds. It divides and is equal to it if and only if the group is cyclic.

Subgroup of false witnesses

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If n is composite, there exists a possibly proper subgroup of , called the "group of false witnesses", comprising the solutions of the equation , the elements which, raised to the power n − 1, are congruent to 1 modulo n.[8] Fermat's Little Theorem states that for n = p a prime, this group consists of all ; thus for n composite, such residues x are "false positives" or "false witnesses" for the primality of n. The number x = 2 is most often used in this basic primality check, and n = 341 = 11 × 31 is notable since , and n = 341 is the smallest composite number for which x = 2 is a false witness to primality. In fact, the false witnesses subgroup for 341 contains 100 elements, and is of index 3 inside the 300-element group .

Examples

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n = 9

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The smallest example with a nontrivial subgroup of false witnesses is 9 = 3 × 3. There are 6 residues coprime to 9: 1, 2, 4, 5, 7, 8. Since 8 is congruent to −1 modulo 9, it follows that 88 is congruent to 1 modulo 9. So 1 and 8 are false positives for the "primality" of 9 (since 9 is not actually prime). These are in fact the only ones, so the subgroup {1,8} is the subgroup of false witnesses. The same argument shows that n − 1 is a "false witness" for any odd composite n.

n = 91

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For n = 91 (= 7 × 13), there are residues coprime to 91, half of them (i.e., 36 of them) are false witnesses of 91, namely 1, 3, 4, 9, 10, 12, 16, 17, 22, 23, 25, 27, 29, 30, 36, 38, 40, 43, 48, 51, 53, 55, 61, 62, 64, 66, 68, 69, 74, 75, 79, 81, 82, 87, 88, and 90, since for these values of x, x90 is congruent to 1 mod 91.

n = 561

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n = 561 (= 3 × 11 × 17) is a Carmichael number, thus s560 is congruent to 1 modulo 561 for any integer s coprime to 561. The subgroup of false witnesses is, in this case, not proper; it is the entire group of multiplicative units modulo 561, which consists of 320 residues.

Examples

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This table shows the cyclic decomposition of and a generating set for n ≤ 128. The decomposition and generating sets are not unique; for example,

(but ). The table below lists the shortest decomposition (among those, the lexicographically first is chosen – this guarantees isomorphic groups are listed with the same decompositions). The generating set is also chosen to be as short as possible, and for n with primitive root, the smallest primitive root modulo n is listed.

For example, take . Then means that the order of the group is 8 (i.e., there are 8 numbers less than 20 and coprime to it); means the order of each element divides 4; that is, the fourth power of any number coprime to 20 is congruent to 1 (mod 20). The set {3,19} generates the group, which means that every element of is of the form 3a × 19b (where a is 0, 1, 2, or 3, because the element 3 has order 4, and similarly b is 0 or 1, because the element 19 has order 2).

Smallest primitive root mod n are (0 if no root exists)

0, 1, 2, 3, 2, 5, 3, 0, 2, 3, 2, 0, 2, 3, 0, 0, 3, 5, 2, 0, 0, 7, 5, 0, 2, 7, 2, 0, 2, 0, 3, 0, 0, 3, 0, 0, 2, 3, 0, 0, 6, 0, 3, 0, 0, 5, 5, 0, 3, 3, 0, 0, 2, 5, 0, 0, 0, 3, 2, 0, 2, 3, 0, 0, 0, 0, 2, 0, 0, 0, 7, 0, 5, 5, 0, 0, 0, 0, 3, 0, 2, 7, 2, 0, 0, 3, 0, 0, 3, 0, ... (sequence A046145 in the OEIS)

Numbers of the elements in a minimal generating set of mod n are

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 3, 1, 2, 1, 2, 2, 1, 1, 3, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 1, 2, 2, 2, 2, 1, 3, 1, 1, 1, 3, 2, 1, 2, 3, 1, 2, ... (sequence A046072 in the OEIS)
Group structure of
Generating set   Generating set   Generating set   Generating set
1 C1 1 1 0 33 C2×C10 20 10 2, 10 65 C4×C12 48 12 2, 12 97 C96 96 96 5
2 C1 1 1 1 34 C16 16 16 3 66 C2×C10 20 10 5, 7 98 C42 42 42 3
3 C2 2 2 2 35 C2×C12 24 12 2, 6 67 C66 66 66 2 99 C2×C30 60 30 2, 5
4 C2 2 2 3 36 C2×C6 12 6 5, 19 68 C2×C16 32 16 3, 67 100 C2×C20 40 20 3, 99
5 C4 4 4 2 37 C36 36 36 2 69 C2×C22 44 22 2, 68 101 C100 100 100 2
6 C2 2 2 5 38 C18 18 18 3 70 C2×C12 24 12 3, 69 102 C2×C16 32 16 5, 101
7 C6 6 6 3 39 C2×C12 24 12 2, 38 71 C70 70 70 7 103 C102 102 102 5
8 C2×C2 4 2 3, 5 40 C2×C2×C4 16 4 3, 11, 39 72 C2×C2×C6 24 6 5, 17, 19 104 C2×C2×C12 48 12 3, 5, 103
9 C6 6 6 2 41 C40 40 40 6 73 C72 72 72 5 105 C2×C2×C12 48 12 2, 29, 41
10 C4 4 4 3 42 C2×C6 12 6 5, 13 74 C36 36 36 5 106 C52 52 52 3
11 C10 10 10 2 43 C42 42 42 3 75 C2×C20 40 20 2, 74 107 C106 106 106 2
12 C2×C2 4 2 5, 7 44 C2×C10 20 10 3, 43 76 C2×C18 36 18 3, 37 108 C2×C18 36 18 5, 107
13 C12 12 12 2 45 C2×C12 24 12 2, 44 77 C2×C30 60 30 2, 76 109 C108 108 108 6
14 C6 6 6 3 46 C22 22 22 5 78 C2×C12 24 12 5, 7 110 C2×C20 40 20 3, 109
15 C2×C4 8 4 2, 14 47 C46 46 46 5 79 C78 78 78 3 111 C2×C36 72 36 2, 110
16 C2×C4 8 4 3, 15 48 C2×C2×C4 16 4 5, 7, 47 80 C2×C4×C4 32 4 3, 7, 79 112 C2×C2×C12 48 12 3, 5, 111
17 C16 16 16 3 49 C42 42 42 3 81 C54 54 54 2 113 C112 112 112 3
18 C6 6 6 5 50 C20 20 20 3 82 C40 40 40 7 114 C2×C18 36 18 5, 37
19 C18 18 18 2 51 C2×C16 32 16 5, 50 83 C82 82 82 2 115 C2×C44 88 44 2, 114
20 C2×C4 8 4 3, 19 52 C2×C12 24 12 7, 51 84 C2×C2×C6 24 6 5, 11, 13 116 C2×C28 56 28 3, 115
21 C2×C6 12 6 2, 20 53 C52 52 52 2 85 C4×C16 64 16 2, 3 117 C6×C12 72 12 2, 17
22 C10 10 10 7 54 C18 18 18 5 86 C42 42 42 3 118 C58 58 58 11
23 C22 22 22 5 55 C2×C20 40 20 2, 21 87 C2×C28 56 28 2, 86 119 C2×C48 96 48 3, 118
24 C2×C2×C2 8 2 5, 7, 13 56 C2×C2×C6 24 6 3, 13, 29 88 C2×C2×C10 40 10 3, 5, 7 120 C2×C2×C2×C4 32 4 7, 11, 19, 29
25 C20 20 20 2 57 C2×C18 36 18 2, 20 89 C88 88 88 3 121 C110 110 110 2
26 C12 12 12 7 58 C28 28 28 3 90 C2×C12 24 12 7, 11 122 C60 60 60 7
27 C18 18 18 2 59 C58 58 58 2 91 C6×C12 72 12 2, 3 123 C2×C40 80 40 7, 83
28 C2×C6 12 6 3, 13 60 C2×C2×C4 16 4 7, 11, 19 92 C2×C22 44 22 3, 91 124 C2×C30 60 30 3, 61
29 C28 28 28 2 61 C60 60 60 2 93 C2×C30 60 30 11, 61 125 C100 100 100 2
30 C2×C4 8 4 7, 11 62 C30 30 30 3 94 C46 46 46 5 126 C6×C6 36 6 5, 13
31 C30 30 30 3 63 C6×C6 36 6 2, 5 95 C2×C36 72 36 2, 94 127 C126 126 126 3
32 C2×C8 16 8 3, 31 64 C2×C16 32 16 3, 63 96 C2×C2×C8 32 8 5, 17, 31 128 C2×C32 64 32 3, 127

See also

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Notes

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References

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Revisions and contributorsEdit on WikipediaRead on Wikipedia
from Grokipedia
The multiplicative group of integers modulo nn, denoted (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times or U(n)U(n), is the finite abelian group consisting of the residue classes modulo nn that are coprime to nn, equipped with the operation of multiplication modulo nn.[1][2] Its order is ϕ(n)\phi(n), where ϕ\phi is Euler's totient function, which counts the number of integers from 1 to n1n-1 that are relatively prime to nn.[1] This group structure arises because the product of two elements coprime to nn remains coprime to nn, multiplication modulo nn is associative, the identity element 1 is coprime to nn, and every such element has a multiplicative inverse modulo nn by Bézout's identity.[2] The group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times is always abelian, meaning the operation is commutative, which follows from the commutativity of integer multiplication.[1][2] Its structure varies with nn: it is cyclic (isomorphic to Z/ϕ(n)Z\mathbb{Z}/\phi(n)\mathbb{Z}) precisely when n=1,2,4,pk,n = 1, 2, 4, p^k, or 2pk2p^k for an odd prime pp and positive integer kk.[1] For other nn, it decomposes into a direct product of cyclic groups; for example, (Z/8Z)×C2×C2(\mathbb{Z}/8\mathbb{Z})^\times \cong C_2 \times C_2, where CmC_m denotes the cyclic group of order mm.[1] In number theory, this group underpins key results such as Euler's theorem, which states that if aa is coprime to nn, then aϕ(n)1(modn)a^{\phi(n)} \equiv 1 \pmod{n}, generalizing Fermat's little theorem for primes.[2] It also plays a central role in cryptography, particularly in the RSA algorithm, where the difficulty of computing discrete logarithms or factorizations in (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times for large composite nn (product of two primes) ensures security for public-key encryption.[3]

Fundamentals

Definition

The multiplicative group of integers modulo $ n $, where $ n $ is a positive integer, is denoted $ (\mathbb{Z}/n\mathbb{Z})^\times $ and consists of the residue classes $ [a] $ with $ 1 \leq a \leq n $ and $ \gcd(a, n) = 1 $, under the operation defined by $ [a] \cdot [b] = [ab \mod n] $.[4] This operation is well-defined on these classes because multiplication and reduction modulo $ n $ preserve the equivalence relation of residues.[5] This structure forms an abelian group. Closure follows from the property that if $ \gcd(a, n) = 1 $ and $ \gcd(b, n) = 1 $, then $ \gcd(ab, n) = 1 $, ensuring the product residue remains coprime to $ n $.[6] The identity element is the residue class $ [1] $, as $ [a] \cdot [1] = [a] $ for any $ [a] $ in the set. Each element $ [a] $ admits a multiplicative inverse $ [x] $ satisfying $ [a] \cdot [x] = [1] $; by Bézout's identity, since $ \gcd(a, n) = 1 $, integers $ x $ and $ y $ exist such that $ ax + ny = 1 $, implying $ ax \equiv 1 \pmod{n} $.[7] Unlike the additive group $ \mathbb{Z}/n\mathbb{Z} $, which comprises all residue classes $ [0], [1], \dots, [n-1] $ under addition modulo $ n $ and includes non-invertible elements like $ [0] $, the multiplicative group $ (\mathbb{Z}/n\mathbb{Z})^\times $ is restricted to the invertible residue classes under multiplication modulo $ n $.[4] The notion arose in Leonhard Euler's 1763 investigation of what is now known as the totient function, which counts the elements of this group.[8]

Elements and Order

The elements of the multiplicative group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times are the equivalence classes of integers modulo nn that are coprime to nn, specifically those represented by integers aa with 1an1 \leq a \leq n and gcd(a,n)=1\gcd(a,n)=1.[9] These residues form the set of units in the ring Z/nZ\mathbb{Z}/n\mathbb{Z}, as each such aa has a multiplicative inverse modulo nn.[10] The order of this group, denoted (Z/nZ)×|(\mathbb{Z}/n\mathbb{Z})^\times|, equals Euler's totient function ϕ(n)\phi(n), which counts the number of integers from 1 to nn that are coprime to nn.[11] For n=1n=1, the group is trivial with the single element [1][1] (equivalently [0][0] modulo 1).[10] The explicit formula for ϕ(n)\phi(n) is
ϕ(n)=npn(11p), \phi(n) = n \prod_{p \mid n} \left(1 - \frac{1}{p}\right),
where the product runs over the distinct prime factors pp of nn.[11] To derive this, first consider the case where n=pkn = p^k is a prime power, with pp prime and k1k \geq 1. Here, the integers from 1 to pkp^k that are not coprime to pkp^k are exactly the multiples of pp, of which there are pk1p^{k-1}. Thus,
ϕ(pk)=pkpk1=pk1(p1). \phi(p^k) = p^k - p^{k-1} = p^{k-1}(p-1).
[10] For a prime pp, this simplifies to ϕ(p)=p1\phi(p) = p-1.[11] For general nn with prime factorization n=p1k1p2k2pmkmn = p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}, ϕ(n)\phi(n) is multiplicative over coprime factors: if gcd(a,b)=1\gcd(a,b)=1, then ϕ(ab)=ϕ(a)ϕ(b)\phi(ab) = \phi(a)\phi(b).[9] This follows from the Chinese Remainder Theorem, which identifies the units modulo abab with pairs of units modulo aa and modulo bb.[10] Therefore,
ϕ(n)=i=1mϕ(piki)=i=1mpiki1(pi1), \phi(n) = \prod_{i=1}^m \phi(p_i^{k_i}) = \prod_{i=1}^m p_i^{k_i-1}(p_i - 1),
yielding the product formula above.[11] For example, if n=pqn = pq with distinct primes pp and qq, then ϕ(pq)=(p1)(q1)\phi(pq) = (p-1)(q-1).[10]

Group Axioms

The set of residue classes modulo nn that are coprime to nn, equipped with multiplication modulo nn, forms a group. This is verified by checking the standard group axioms: closure, associativity, identity element, and existence of inverses. Additionally, the group is abelian due to the commutativity of multiplication. To establish closure, suppose [a][a] and [b][b] are elements of the group, meaning gcd(a,n)=1\gcd(a, n) = 1 and gcd(b,n)=1\gcd(b, n) = 1. The product is [a][b]=[abmodn][a][b] = [ab \mod n]. It must be shown that gcd(abmodn,n)=1\gcd(ab \mod n, n) = 1. Note that gcd(abmodn,n)=gcd(ab,n)\gcd(ab \mod n, n) = \gcd(ab, n), since ab=qn+(abmodn)ab = qn + (ab \mod n) for some integer qq, and any common divisor of abmodnab \mod n and nn divides abab. Now, gcd(ab,n)=1\gcd(ab, n) = 1 because if a prime pp divides both abab and nn, then pp divides aa or bb, but gcd(a,n)=1\gcd(a, n) = 1 implies pp does not divide aa, and similarly for bb, yielding a contradiction unless no such pp exists. Thus, the product is coprime to nn and belongs to the set.[12] Associativity follows from the associativity of integer multiplication: for [a],[b],[c][a], [b], [c] in the set, ([a][b])[c]=[(abmodn)cmodn]=[a(bcmodn)modn]=[a]([b][c])([a][b])[c] = [(ab \mod n) c \mod n] = [a (bc \mod n) \mod n] = [a]([b][c]), as the modulo operation preserves the associative property of multiplication in Z\mathbb{Z}.[13] The identity element is [1][1], since for any [a][a] in the set, [a][1]=[a1modn]=[a][a][1] = [a \cdot 1 \mod n] = [a], and similarly [1][a]=[a][1][a] = [a].[14] For inverses, given [a][a] with gcd(a,n)=1\gcd(a, n) = 1, Bézout's identity guarantees integers x,yx, y such that ax+ny=1a x + n y = 1, so ax1modna x \equiv 1 \mod n, meaning [x][x] satisfies [a][x]=[1][a][x] = [1] and is thus the inverse of [a][a]. This inverse can be computed explicitly using the extended Euclidean algorithm.[15] The group is abelian because multiplication of integers is commutative: abbamodnab \equiv ba \mod n, so [a][b]=[b][a][a][b] = [b][a] for all elements.[13]

Properties

Notation

The multiplicative group of integers modulo nn, where nn is a positive integer greater than 1, is commonly denoted by (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, representing the group of units in the quotient ring Z/nZ\mathbb{Z}/n\mathbb{Z}. This notation emphasizes its structure as the set of residue classes coprime to nn under multiplication modulo nn. An alternative and widely used symbol is U(n)U(n), which highlights the group's role as the units of the ring Z/nZ\mathbb{Z}/n\mathbb{Z}.[16] Elements of the group are typically represented as equivalence classes, such as a\overline{a} for the residue class of an integer aa modulo nn, where 1a<n1 \leq a < n and gcd(a,n)=1\gcd(a, n) = 1, or alternatively as [a][a].[17] The group operation is multiplication modulo nn, expressed as ab=abmodn\overline{a} \cdot \overline{b} = \overline{ab \mod n}, ensuring the result remains within the residue classes coprime to nn.[1] The order of the group, or the number of its elements, is denoted by Euler's totient function ϕ(n)\phi(n), which counts the integers up to nn that are coprime to nn.[11] In some older mathematical texts, the group may be denoted by Γ(n)\Gamma(n), though this usage is less common in modern literature.[18] As the unit group of the ring Z/nZ\mathbb{Z}/n\mathbb{Z}, (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times consists precisely of those elements that possess multiplicative inverses modulo nn.

Cardinality

The multiplicative group of integers modulo nn, denoted (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, has cardinality ϕ(n)\phi(n), where ϕ\phi denotes Euler's totient function, which counts the number of integers from 1 to n1n-1 that are coprime to nn.[11] The function ϕ\phi is multiplicative, so that if gcd(m,n)=1\gcd(m,n)=1, then ϕ(mn)=ϕ(m)ϕ(n)\phi(mn)=\phi(m)\phi(n).[11] Additionally, ϕ(n)\phi(n) is even for all n>2n>2.[11] By Lagrange's theorem applied to the finite group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, the order of any subgroup divides ϕ(n)\phi(n).[19] In particular, for any element a(Z/nZ)×a \in (\mathbb{Z}/n\mathbb{Z})^\times, the order of aa divides ϕ(n)\phi(n). This implies Euler's theorem: if gcd(a,n)=1\gcd(a,n)=1, then aϕ(n)1(modn)a^{\phi(n)} \equiv 1 \pmod{n}.[19] To sketch the proof of Euler's theorem using Lagrange's result, consider the cyclic subgroup HH generated by aa, which has order dd equal to the multiplicative order of aa modulo nn. By Lagrange's theorem, dd divides ϕ(n)\phi(n), so ad1(modn)a^d \equiv 1 \pmod{n} and thus aϕ(n)=(ad)ϕ(n)/d1(modn)a^{\phi(n)} = (a^d)^{\phi(n)/d} \equiv 1 \pmod{n}.[19] Quantitative bounds on the size of the group are also known; in particular, there exists a constant c>0c>0 such that ϕ(n)>cn/loglogn\phi(n) > c n / \log \log n for all sufficiently large nn.[20]

Isomorphisms via Chinese Remainder Theorem

A fundamental result concerning the structure of the multiplicative group of integers modulo nn arises from the Chinese Remainder Theorem when nn factors into coprime parts. Specifically, if n=n1n2n = n_1 n_2 where gcd(n1,n2)=1\gcd(n_1, n_2) = 1, then there is a group isomorphism
(Z/nZ)×(Z/n1Z)××(Z/n2Z)× (\mathbb{Z}/n\mathbb{Z})^\times \cong (\mathbb{Z}/n_1\mathbb{Z})^\times \times (\mathbb{Z}/n_2\mathbb{Z})^\times
given explicitly by the map sending the residue class [amodn][a \mod n] to the pair ([amodn1],[amodn2])([a \mod n_1], [a \mod n_2]). This isomorphism preserves the multiplicative operation, as multiplication in the product group corresponds to componentwise multiplication modulo n1n_1 and n2n_2, which aligns with reduction modulo nn via the Chinese Remainder Theorem.[21] The proof of this isomorphism follows from the ring-theoretic version of the Chinese Remainder Theorem, which establishes a ring isomorphism Z/nZZ/n1Z×Z/n2Z\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/n_1\mathbb{Z} \times \mathbb{Z}/n_2\mathbb{Z}. Since units in a product ring are precisely the pairs of units from each factor, the induced map on unit groups is a group isomorphism. To verify, the map is a homomorphism because it respects multiplication and reduction modulo the factors; it is injective since equal images imply congruence modulo both n1n_1 and n2n_2, hence modulo nn; and it is surjective by constructing preimages using the existence of solutions to simultaneous congruences for units.[21] This result generalizes straightforwardly to the prime power factorization of nn. If n=i=1rpikin = \prod_{i=1}^r p_i^{k_i} where the pip_i are distinct primes and ki1k_i \geq 1, then by iterated application of the two-factor case (via induction on the number of factors), the group decomposes as
(Z/nZ)×i=1r(Z/pikiZ)×. (\mathbb{Z}/n\mathbb{Z})^\times \cong \prod_{i=1}^r (\mathbb{Z}/p_i^{k_i}\mathbb{Z})^\times.
The explicit isomorphism sends [amodn][a \mod n] to the tuple ([amodpiki]i=1r)([a \mod p_i^{k_i}]_{i=1}^r), leveraging the coprimality of the prime power moduli.[22] As a consequence, the structure of (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times for composite nn is fully determined by the structures of the unit groups modulo prime powers, reducing the problem to analyzing those simpler cases. This decomposition is a cornerstone for classifying the group across all nn.[22]

Structure

Cyclic Case

The multiplicative group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times is cyclic if and only if n=1,2,4,pk,n = 1, 2, 4, p^k, or 2pk2p^k where pp is an odd prime and k1k \geq 1.[23] Under these conditions, the group admits generators known as primitive roots modulo nn; an integer aa coprime to nn is a primitive root if its multiplicative order modulo nn equals ϕ(n)\phi(n), the Euler totient function.[24] The existence of such generators ensures that every element coprime to nn can be expressed as a power of the primitive root modulo nn.[23] When (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times is cyclic, it is isomorphic to the additive cyclic group Z/ϕ(n)Z\mathbb{Z}/\phi(n)\mathbb{Z} of order ϕ(n)\phi(n).[25] A key structural property is that the group contains exactly one subgroup of order dd for each positive divisor dd of ϕ(n)\phi(n); these subgroups consist of the elements whose orders divide dd.[25] The number of primitive roots modulo nn is precisely ϕ(ϕ(n))\phi(\phi(n)), reflecting the count of generators in a cyclic group of that order.[23] To identify a primitive root modulo nn, factorize ϕ(n)\phi(n) into its prime factors q1,q2,,qrq_1, q_2, \dots, q_r, and test candidate integers gg (starting from small values like 2 or 3, ensuring gcd(g,n)=1\gcd(g, n) = 1) by checking that gϕ(n)/qi≢1(modn)g^{\phi(n)/q_i} \not\equiv 1 \pmod{n} for each i=1,,ri = 1, \dots, r.[24] If the condition holds for all prime factors, then the order of gg is exactly ϕ(n)\phi(n), confirming it as a primitive root; otherwise, proceed to the next candidate.[24] This method leverages the fact that the order must divide ϕ(n)\phi(n) but not any proper divisor obtained by removing one prime factor.

Powers of Odd Primes

The multiplicative group (Z/pkZ)×(\mathbb{Z}/p^k\mathbb{Z})^\times, where pp is an odd prime and k1k \geq 1, is cyclic of order ϕ(pk)=pk1(p1)\phi(p^k) = p^{k-1}(p-1).[26] This order counts the integers from 1 to pk1p^k - 1 that are coprime to pkp^k, excluding multiples of pp.[27] As a cyclic group of order ϕ(pk)\phi(p^k), (Z/pkZ)×(\mathbb{Z}/p^k\mathbb{Z})^\times is isomorphic to the cyclic group Z/ϕ(pk)Z\mathbb{Z}/\phi(p^k)\mathbb{Z}.[25] Primitive roots modulo pkp^k exist, meaning there are generators gg such that the powers of gg modulo pkp^k produce all units. Specifically, if gg is a primitive root modulo pp, it can be lifted to a primitive root modulo pkp^k using Hensel's lemma applied to the polynomial f(x)=xϕ(pk)1=0f(x) = x^{\phi(p^k)} - 1 = 0, ensuring the lift preserves the order since the derivative condition f(g)≢0(modp)f'(g) \not\equiv 0 \pmod{p} holds.[28] The subgroup structure follows from cyclicity: for each positive divisor dd of ϕ(pk)\phi(p^k), there is a unique cyclic subgroup of order dd, generated by gϕ(pk)/dg^{\phi(p^k)/d} for any generator gg.[25] In particular, the unique subgroup of order pk1p^{k-1} consists of elements congruent to 1 modulo pp, i.e., of the form 1+mp1 + m p for integers mm modulo pk1p^{k-1}, and this subgroup is cyclic, generated by 1+p1 + p. Such lifts ensure the full group structure is captured by iteratively adjusting primitive roots from lower powers, as in the construction where a primitive root modulo pkp^k yields pp primitive roots modulo pk+1p^{k+1} by adding multiples of pkp^k.[29]

Powers of 2

For k=1k = 1, the group (Z/2Z)×(\mathbb{Z}/2\mathbb{Z})^\times is the trivial group consisting of the single element 11.[22] For k=2k = 2, the group (Z/4Z)×={1,3}(\mathbb{Z}/4\mathbb{Z})^\times = \{1, 3\} is cyclic of order 22, generated by 33.[18] For k3k \geq 3, the group (Z/2kZ)×(\mathbb{Z}/2^k\mathbb{Z})^\times is isomorphic to Z/2Z×Z/2k2Z\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2^{k-2}\mathbb{Z}.[22] This isomorphism is explicit, with the group generated by 1-1 (of order 22) and 55 (of order 2k22^{k-2}).[18] Consequently, the group is non-cyclic, as its maximum element order is 2k22^{k-2}, which is strictly less than ϕ(2k)=2k1\phi(2^k) = 2^{k-1}.[22] The cyclic subgroup of order 2k22^{k-2} is generated by 55 and consists precisely of the elements congruent to 1(mod4)1 \pmod{4}.[18]

General Composite Case

For a general composite integer n>1n > 1 with prime factorization n=2ki=1rpiein = 2^k \prod_{i=1}^r p_i^{e_i}, where the pip_i are distinct odd primes and k0k \geq 0, ei1e_i \geq 1, the multiplicative group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times is isomorphic to the direct product (Z/2kZ)××i=1r(Z/pieiZ)×(\mathbb{Z}/2^k \mathbb{Z})^\times \times \prod_{i=1}^r (\mathbb{Z}/p_i^{e_i} \mathbb{Z})^\times by the Chinese Remainder Theorem applied to the coprime factors of nn.[21] The structures of the prime power components (Z/peZ)×(\mathbb{Z}/p^e \mathbb{Z})^\times are known from the cases of powers of 2 and odd primes.[21] This decomposition yields non-cyclic groups in many composite cases. For example, when n=15=35n = 15 = 3 \cdot 5, (Z/15Z)×(Z/3Z)××(Z/5Z)×Z/2Z×Z/4Z(\mathbb{Z}/15\mathbb{Z})^\times \cong (\mathbb{Z}/3\mathbb{Z})^\times \times (\mathbb{Z}/5\mathbb{Z})^\times \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}, which is non-cyclic of order ϕ(15)=8\phi(15) = 8.[21] Similarly, for n=24=83n = 24 = 8 \cdot 3, (Z/24Z)×(Z/8Z)××(Z/3Z)×Z/2Z×Z/2Z×Z/2Z(\mathbb{Z}/24\mathbb{Z})^\times \cong (\mathbb{Z}/8\mathbb{Z})^\times \times (\mathbb{Z}/3\mathbb{Z})^\times \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}, a non-cyclic elementary abelian group of order ϕ(24)=8\phi(24) = 8.[21] The exponent of (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, the least common multiple of the orders of its elements, is the least common multiple of the exponents of the component groups (Z/peZ)×(\mathbb{Z}/p^e \mathbb{Z})^\times.[21] This exponent equals the Carmichael function λ(n)\lambda(n), defined as λ(n)=lcm(λ(2k),λ(p1e1),,λ(prer))\lambda(n) = \mathrm{lcm}(\lambda(2^k), \lambda(p_1^{e_1}), \dots, \lambda(p_r^{e_r})), where λ\lambda on prime powers matches the respective group exponents.[30] The group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times fails to be cyclic precisely when nn has two or more distinct odd prime factors, or when nn is divisible by 8 and by an odd prime, or when n=2kn = 2^k for k3k \geq 3; it is cyclic only for n=1,2,4,pe,n = 1, 2, 4, p^e, or 2pe2p^e with pp an odd prime and e1e \geq 1.[31]

Special Subgroups

Quadratic Residues

In the multiplicative group of integers modulo nn, denoted (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, the quadratic residues form the subgroup consisting of all squares of elements in the group. Formally, this subgroup is {[a2][a](Z/nZ)×}\{ [a^2] \mid [a] \in (\mathbb{Z}/n\mathbb{Z})^\times \}, where [][ \cdot ] denotes the congruence class modulo nn. The squaring map xx2x \mapsto x^2 is a group endomorphism of (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times whose image is this subgroup of quadratic residues and whose kernel is the set of units xx satisfying x21(modn)x^2 \equiv 1 \pmod{n}. For n>2n > 2, ϕ(n)\phi(n) is even and this kernel always contains {[1],[1]}\{ [1], [-1] \}, but its size is 2 only in specific cases such as odd prime powers or n=4n=4; in general, the size (and thus the index of the quadratic residues subgroup) is 2r2^r where rr is the number of cyclic factors of even order in the primary decomposition of the group.[22] For an odd prime pp, the subgroup of quadratic residues in (Z/pZ)×(\mathbb{Z}/p\mathbb{Z})^\times is cyclic of order (p1)/2(p-1)/2. This follows from the cyclicity of (Z/pZ)×(\mathbb{Z}/p\mathbb{Z})^\times and the fact that the quadratic residues are precisely the even powers of a primitive root modulo pp. The Legendre symbol provides a means to detect membership in this subgroup: for an integer aa not divisible by pp, (ap)=a(p1)/2(modp)\left( \frac{a}{p} \right) = a^{(p-1)/2} \pmod{p}, which equals 1 if aa is a quadratic residue modulo pp, -1 if it is a non-residue, and 0 if pp divides aa.[32][33] For powers of 2 with exponent k3k \geq 3, the subgroup of quadratic residues has index 4; for example, modulo 8 it is {1}\{1\}, and modulo 16 it is {1,9}\{1, 9\}.[22] For composite n>2n > 2, the structure of the subgroup of quadratic residues is determined via the Chinese Remainder Theorem, as (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times decomposes into a direct product of the groups (Z/pkZ)×(\mathbb{Z}/p^k\mathbb{Z})^\times for the prime power factors pkp^k of nn. An element is a quadratic residue modulo nn if and only if it is a quadratic residue modulo each such prime power. For an odd prime power pkp^k with k1k \geq 1, the subgroup has index 2 and order ϕ(pk)/2=pk1(p1)/2\phi(p^k)/2 = p^{k-1}(p-1)/2. The overall index is the product of the local indices.[34]

False Witnesses

In the context of primality testing for a composite integer n>1n > 1, a false witness, also known as a Fermat liar, is an integer aa coprime to nn such that an11(modn)a^{n-1} \equiv 1 \pmod{n}.[35] This condition mimics the conclusion of Fermat's Little Theorem, which states that if nn is prime, then an11(modn)a^{n-1} \equiv 1 \pmod{n} for all aa coprime to nn; thus, elements aa where an1≢1(modn)a^{n-1} \not\equiv 1 \pmod{n} serve as witnesses to the compositeness of nn.[35] Similarly, in the Miller-Rabin primality test, false witnesses (or strong liars) are bases aa that satisfy the stronger conditions derived from writing n1=2sdn-1 = 2^s d with dd odd, namely ad1(modn)a^d \equiv 1 \pmod{n} or a2rd1(modn)a^{2^r d} \equiv -1 \pmod{n} for some 0r<s0 \leq r < s.[36] The set of Fermat false witnesses forms a proper subgroup of the multiplicative group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, consisting of all units aa modulo nn satisfying an11(modn)a^{n-1} \equiv 1 \pmod{n}.[35] This subgroup contains the identity element 1, since 1n11(modn)1^{n-1} \equiv 1 \pmod{n}, and is closed under multiplication: if an11(modn)a^{n-1} \equiv 1 \pmod{n} and bn11(modn)b^{n-1} \equiv 1 \pmod{n}, then (ab)n1=an1bn1111(modn)(ab)^{n-1} = a^{n-1} b^{n-1} \equiv 1 \cdot 1 \equiv 1 \pmod{n}.[35] It is also closed under inversion, as the finite order of elements in (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times ensures that if the order of aa divides n1n-1, then the order of a1a^{-1} does as well.[35] The order of this subgroup divides ϕ(n)\phi(n), the order of (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, but is strictly less than ϕ(n)\phi(n) unless nn is a Carmichael number, in which case every unit is a false witness and the subgroup coincides with the full group.[35] In contrast, the set of Miller-Rabin false witnesses does not necessarily form a subgroup of (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, though it is contained within a proper subgroup of index at most 4, ensuring that at least three-quarters of the units are witnesses for odd composite nn.[36] This structural distinction highlights the enhanced reliability of the Miller-Rabin test over the basic Fermat test in probabilistic primality verification.[36]

Examples of False Witnesses

One concrete example of false witnesses occurs for n=9=32n = 9 = 3^2, where the multiplicative group (Z/9Z)×={1,2,4,5,7,8}(\mathbb{Z}/9\mathbb{Z})^\times = \{1, 2, 4, 5, 7, 8\} has order ϕ(9)=6\phi(9) = 6. For the Fermat primality test, false witnesses are bases aa coprime to 9 satisfying a81(mod9)a^{8} \equiv 1 \pmod{9}. Direct computation shows this holds for a=1a = 1 (trivially, as 18=11^{8} = 1) and a=8a = 8 (since 821(mod9)8^{2} \equiv 1 \pmod{9} and 88=(82)414=1(mod9)8^{8} = (8^{2})^{4} \equiv 1^{4} = 1 \pmod{9}), but fails for the others (e.g., 28=2564(mod9)2^{8} = 256 \equiv 4 \pmod{9}, 487(mod9)4^{8} \equiv 7 \pmod{9}).[35] Thus, there are 2 false witnesses out of 6 possible bases, a proportion of 1/31/3.[37] Another example is n=91=7×13n = 91 = 7 \times 13, with group order ϕ(91)=72\phi(91) = 72. False witnesses satisfy a901(mod91)a^{90} \equiv 1 \pmod{91} for aa coprime to 91. Base 3 is one such false witness, as its order in (Z/91Z)×(\mathbb{Z}/91\mathbb{Z})^\times is 6 (lcm of orders 6 modulo 7 and 3 modulo 13), dividing 90, so 390=(36)15115=1(mod91)3^{90} = (3^{6})^{15} \equiv 1^{15} = 1 \pmod{91}.[38] The set of all false witnesses forms a subgroup of size 36 (computed via CRT: all 6 elements modulo 7 and the 6 elements of order dividing 6 modulo 13 satisfy the condition, yielding 6×6=366 \times 6 = 36).[39] This gives a proportion of 36/72=1/236/72 = 1/2 false witnesses.[40] A more extreme case is the Carmichael number n=561=3×11×17n = 561 = 3 \times 11 \times 17, with ϕ(561)=320\phi(561) = 320. As the smallest Carmichael number, it satisfies a5601(mod561)a^{560} \equiv 1 \pmod{561} for every aa coprime to 561, making all 320 elements of the group false witnesses relative to the Fermat test (verified by the defining property that λ(561)\lambda(561) divides 560, where λ\lambda is the Carmichael function). The proportion is thus 1, with the entire group consisting of false witnesses.[35]

Applications

Primality Testing

The Fermat primality test leverages the structure of the multiplicative group of integers modulo n, denoted (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, to assess whether n is prime. For a candidate prime n > 2, select a random base a with 1<a<n1 < a < n and gcd(a,n)=1\gcd(a, n) = 1, then verify if an11(modn)a^{n-1} \equiv 1 \pmod{n}. If the congruence fails, n is composite. This test is grounded in Fermat's Little Theorem, which states that if n is prime, then the order of every element in the group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times divides n-1, the group's order, ensuring the congruence holds for all such a.[35] However, the test can err on composite n that are Fermat pseudoprimes to base a, where the congruence holds despite compositeness. The test fails deterministically for Carmichael numbers, which are square-free composite integers n such that the exponent of (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times—the least common multiple of the orders of its elements—divides n-1. For these n, an11(modn)a^{n-1} \equiv 1 \pmod{n} holds for all a coprime to n, mimicking prime behavior and causing the Fermat test to incorrectly classify them as probable primes.[41] This vulnerability arises because the group's exponent dividing n-1 implies every element's order divides n-1, just as in the prime case, though the group is not cyclic. To address these shortcomings, the Miller-Rabin test provides a stronger probabilistic primality test, also exploiting properties of (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times. Write n-1 = 2sd2^s \cdot d with d odd, then for a random base a coprime to n, check if either ad1(modn)a^d \equiv 1 \pmod{n} or a2rd1(modn)a^{2^r d} \equiv -1 \pmod{n} for some 0r<s0 \leq r < s. If neither holds, a is a witness to compositeness. For prime n, the group is cyclic of order n-1, ensuring every a satisfies the conditions due to the existence of elements of order dividing n-1 with specific square root behaviors. If n is composite, at least three-fourths of bases are witnesses, yielding an error probability less than 1/41/4 per trial; multiple independent trials reduce this exponentially.[36] These tests rely on the cyclic nature of (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times for prime n, where a generator exists with order exactly n-1, guaranteeing the required congruences without the pitfalls seen in non-cyclic composite groups. Deterministic variants eliminate randomness by testing a fixed set of small bases; for example, bases 2, 3, 5, 7, 11, 13, 17, 19, and 23 suffice for all odd n < 3,825,123,056,546,413,051 (about 62 bits).[42]

Cryptography

The Diffie-Hellman key exchange protocol utilizes the multiplicative group of integers modulo a prime pp, specifically cyclic subgroups generated by a primitive root g(Z/pZ)×g \in (\mathbb{Z}/p\mathbb{Z})^\times, to enable secure key agreement between two parties over an insecure channel.[43] In this scheme, one party computes gamodpg^a \mod p and sends it to the other, who responds with gbmodpg^b \mod p; the shared secret is then derived as (ga)b=(gb)a=gabmodp(g^a)^b = (g^b)^a = g^{ab} \mod p, without either party revealing their private exponents aa or bb.[43] The security of Diffie-Hellman relies on the hardness of the discrete logarithm problem (DLP) in the cyclic group (Z/pZ)×(\mathbb{Z}/p\mathbb{Z})^\times for large primes pp, where given gg and h=gxmodph = g^x \mod p, computing the exponent xx (the discrete log) is computationally infeasible with current algorithms.[43] This intractability ensures that an eavesdropper cannot efficiently recover the shared secret gabmodpg^{ab} \mod p from the public values g,gamodp,g, g^a \mod p, and gbmodpg^b \mod p, as solving the DLP would be required to extract aa or bb.[44] The RSA cryptosystem employs the multiplicative group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times where n=pqn = pq for distinct large primes pp and qq, with the public exponent ee chosen such that gcd(e,ϕ(n))=1\gcd(e, \phi(n)) = 1, allowing encryption as c=memodnc = m^e \mod n for message mm.[45] Decryption uses the private exponent d=e1modϕ(n)d = e^{-1} \mod \phi(n), recovering m=cdmodnm = c^d \mod n via Euler's theorem, since mϕ(n)1modnm^{\phi(n)} \equiv 1 \mod n for gcd(m,n)=1\gcd(m, n) = 1.[45] RSA's security stems from the difficulty of factoring nn to compute ϕ(n)=(p1)(q1)\phi(n) = (p-1)(q-1), as knowledge of ϕ(n)\phi(n) enables finding dd and thus breaking the system, while factoring large semiprimes remains intractable.[46]

Further Examples

For the prime modulus n=5n=5, the multiplicative group (Z/5Z)×(\mathbb{Z}/5\mathbb{Z})^\times consists of the elements {1,2,3,4}\{1, 2, 3, 4\} under multiplication modulo 5. This group is cyclic of order 4, generated by 2, as the powers are 2122^1 \equiv 2, 2242^2 \equiv 4, 2332^3 \equiv 3, 241(mod5)2^4 \equiv 1 \pmod{5}. The orders of the elements are: 1 has order 1, 2 has order 4, 3 has order 4, and 4 has order 2.[22] For the composite modulus n=15=35n=15=3 \cdot 5, the group (Z/15Z)×(\mathbb{Z}/15\mathbb{Z})^\times has elements {1,2,4,7,8,11,13,14}\{1, 2, 4, 7, 8, 11, 13, 14\}, which is isomorphic to Z2×Z4\mathbb{Z}_2 \times \mathbb{Z}_4 and has order 8. The maximum order of any element is 4, for example, the order of 2 is 4 since 241(mod15)2^4 \equiv 1 \pmod{15} and no smaller positive exponent works.[22] For the power-of-2 modulus n=8=23n=8=2^3, the group (Z/8Z)×(\mathbb{Z}/8\mathbb{Z})^\times consists of the elements {1,3,5,7}\{1, 3, 5, 7\} under multiplication modulo 8, isomorphic to Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2 with order 4. All non-identity elements have order at most 2, as 3213^2 \equiv 1, 5215^2 \equiv 1, and 721(mod8)7^2 \equiv 1 \pmod{8}.[22] The following table summarizes Euler's totient function ϕ(n)\phi(n), the isomorphism type of (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^\times, and an example generator (when the group is cyclic) for n=1n=1 to 20, based on the standard classification of these groups.[22]
nnϕ(n)\phi(n)StructureExample Generator
11Trivial
21Trivial
32Z2\mathbb{Z}_22
42Z2\mathbb{Z}_23
54Z4\mathbb{Z}_42
62Z2\mathbb{Z}_25
76Z6\mathbb{Z}_63
84Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2
96Z6\mathbb{Z}_62
104Z4\mathbb{Z}_43
1110Z10\mathbb{Z}_{10}2
124Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2
1312Z12\mathbb{Z}_{12}2
146Z6\mathbb{Z}_65
158Z2×Z4\mathbb{Z}_2 \times \mathbb{Z}_4
168Z2×Z4\mathbb{Z}_2 \times \mathbb{Z}_4
1716Z16\mathbb{Z}_{16}3
186Z6\mathbb{Z}_65
1918Z18\mathbb{Z}_{18}2
208Z2×Z4\mathbb{Z}_2 \times \mathbb{Z}_4

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