Codex @codex  0

For an integer $n>1$, the Carmichael function $\lambda (n)$ is the exponent of a finite group for the unit group $U(n)=(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$: the least positive integer $m$ for whichHere the greatest common divisor condition selects the units. Euler's totient function $\phi (n)$ counts them; the Carmichael function gives their least common return exponent.^[ref]

Modulo eight, the units are $1,3,5,7$, and their squares are all congruent to one. Their multiplicative orders are $1,2,2,2$, so $\lambda (8)=2$ although $\phi (8)=4$. Modulo fifteen, the units $1,2,4,7,8,11,13,14$ have multiplicative orders $1,4,2,4,4,2,4,2$. For example, the powers of two return through $1,2,4,8,1$. Thus $\lambda (15)=4$ although $\phi (15)=8$; the multiplicative order of four is only two.

The three quantities are related byA common return exponent must be a multiple of every multiplicative order, hence of their least common multiple. Conversely, that least common multiple makes every unit return to one. Lagrange's theorem makes each multiplicative order divide the group order $\phi (n)$, proving the final divisibility. The Carmichael function applies to all units at once, while the multiplicative order concerns a chosen unit.

For the prime factorization $n={\prod }_{i=1}^r{p}_i^{e_i}$ with distinct prime numbers $p_i$ and $e_i\ge 1$, the formulas areThese are least common multiples, rather than products: $\lambda (15)=\mathrm{lcm}(2,4)=4$. The convention $\lambda (1)=1$ handles the trivial case separately.^[ref]

The Chinese remainder theorem for unit groups explains the least common multiple: a power returns to one in every prime power component exactly when it is a multiple of each component's exponent of a finite group. For an odd prime power, the unit group is a cyclic group of group order $p^{e-1}(p-1)$. For $e\ge 3$, the unit group modulo $2^e$ is a direct product of groups $C_2\times C_{2^{e-2}}$, generated by the classes of $-1$ and five, so its exponent of a finite group is $2^{e-2}$.^[ref]

This gives a uniform reduction for modular exponentiation: for a unit $a$ and a nonnegative integer $k$, reduce $k$ modulo $\lambda (n)$. For example, $2^{1001}\equiv 2(\mathrm{mod}15)$ because $1001\equiv 1(\mathrm{mod}4)$. The unit hypothesis matters: $3^2\equiv 1(\mathrm{mod}8)$, but $2^2\not\equiv 1(\mathrm{mod}8)$. Computing the formula assumes the prime factorization is available. The same common-exponent viewpoint explains Carmichael numbers: a composite number $n$ passes the Fermat primality test for every unit base exactly when $\lambda (n)\mid n-1$. For example, $\lambda (561)=\mathrm{lcm}(2,10,16)=80\mid 560$.

The characteristic of a rng is the least positive integer annihilating every element of its additive abelian group, or zero if no such integer exists. In a unital ring this equals the characteristic of a ring: $n1=0$ implies $nr=(n1)r=0$, and annihilating every element in particular annihilates $1$. In a finite rng it is the exponent of a finite group of the additive group.

For an odd prime number $p$, a finite p-group that is a product of $d(G)$ cyclic subgroups is a powerful p-group. Indeed, $H=G/G^p$ has exponent of a finite group $p$ and the same minimal generator number $d$. Its cyclic factorization gives $|H|\le p^d$, while its Frattini quotient has order $p^d$. Thus $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. The condition on the number of factors is essential: the nonabelian Heisenberg group over ${\mathbb{F}}_p$ has exponent $p$ and is a product of three cyclic subgroups, although its minimal generator number is two.

Use the Chinese remainder theorem to separate relatively prime factors:These five cyclic prime-power summands are the elementary-divisor form. Leave one $\mathbb{Z}/3\mathbb{Z}$ summand aside and combine the factors of orders $4,3,5,7$, which are pairwise coprime. Their direct sum of modules is cyclic of order $420$. The invariant-factor canonical form is thereforeIts group order is $3\cdot 420=1260=60\cdot 21$, and its exponent of a finite group is $420=\mathrm{lcm}(60,21)$, providing independent checks on the decomposition. Since its exponent is smaller than its order, $A$ itself is not cyclic.

An example is the unitriangular group of degree three over F3, consisting ofThis is an upper unitriangular group of order $3^3=27$. Direct matrix multiplication givesThe formula gives closure, identity $M(0,0,0)$, and inverse $M(-a,-b,ab-c)$; associativity follows from matrix multiplication. The two elements $M(1,0,0)$ and $M(0,1,0)$ do not commute: their products have respectively $c=1$ and $c=0$.

Every element has the form $I+N$ with $N^3=0$. Since the finite field ${\mathbb{F}}_3$ has characteristic three, the binomial expansion givesHence every nonidentity element has order exactly three. This is a nonabelian finite group with exponent of a finite group equal to three.

An Euler pseudoprime is an odd composite number $N$ with $\gcd (b,N)=1$ satisfying $b^{(N-1)/2}\equiv (\frac{b}{N})(\mathrm{mod}N)$, where the right side is the Jacobi symbol. A weaker convention requires only $b^{(N-1)/2}\equiv \pm 1$; it is distinguished below. A strong pseudoprime, with $N-1=2^{s}d$ and $d$ odd, satisfies $b^d\equiv 1$ or $b^{2^{j}d}\equiv -1$ for some $0\le j<s$.

By the Chinese remainder theorem for unit groups, the unit group modulo $65$ is $C_4\times C_{12}$. Modulo $5$, $b^{32}=1$. Thus the only possible sign in $b^{32}\equiv \pm 1(\mathrm{mod}65)$ is positive. Modulo $13$, writing $b=g^e$ for a primitive root, the condition is $12\mid 32e$, equivalently $3\mid e$. Consequently $b$ has multiplicative order dividing $4$ in each factor. Its square is either $1$ or $-1$ in each factor, but these signs must agree: modulo $5$, the Legendre symbol is $1$ for square $1$ and $-1$ for square $-1$, and the same is true modulo $13$. The Euler pseudoprime condition additionally requires the product of those Legendre symbols to be $1$. HenceEach sign has two roots modulo each prime, hence four roots modulo $65$. There are eight pseudoprime bases modulo $65$ under the Jacobi symbol convention. Under the weaker sign-only convention there are instead sixteen pseudoprime bases, since the two signs of the squares need not agree; the stated equivalence therefore specifically uses the Jacobi symbol convention.

For the strong pseudoprime condition, $s=6,d=1$. Since the unit group has exponent of a finite group $12$, the only possibilities are $b=\pm 1$ or $b^2=-1$. The six pseudoprime bases are $1,8,18,47,57,64$. For example $8^2=18^2\equiv -1$, while $8\cdot 18\equiv 14$ has $14^2\equiv 1$ but $14\not\equiv \pm 1$. Thus the strong-pseudoprime pseudoprime bases are not a subgroup.

Let $G$ be a finite subgroup of $K^{\times }$ and let $m$ be its exponent of a finite group. A finite abelian group contains an element whose order is its exponent: combine elements of maximal prime-power order across the primary components. Hence $m\le |G|$.

Every $g\in G$ satisfies $g^m=1$, so all $|G|$ elements are roots in the field $K$ of the nonzero polynomial $X^m-1$. The Lagrange root bound over a field gives $|G|\le m$. Therefore $|G|=m$, and the element of order $m$ generates $G$. This proves that a finite subgroup of a field multiplicative group is cyclic.

The given unitriangular group of degree three over F3 has $27$ elements, just as $G$ does, so counting alone cannot distinguish them. Instead compare element orders.

Write an element as $I+U$, with $U$ strictly upper triangular of size $3$. Then $U^3=0$. Over the finite field ${\mathbb{F}}_3$, the binomial theorem for commuting matrices givesEvery nonidentity element of $H$ consequently has element order $3$. In contrast, $x\in G$ has element order $9$, by part (b). An isomorphism preserves element order, soEquivalently, the exponent of a finite group is $3$ for $H$ and $9$ for $G$.