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ExampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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The unit group modulo one hundred is isomorphic to C_20 times C_2

Example

In the unit group U(100)U(100), the class of 33 has order 2020 and the class of 1-1 has order 22. Their subgroups form an internal direct product, so U(100)C20×C2,U(100)\cong C_{20}\times C_2, with invariant factors 2202\mid20.

Facts & Assumptions

Given: The objects and hypotheses in the example.

[L1]

Let n1n\ge1 be an integer. Multiplication makes Z/n\mathbb Z/n a commutative monoid with identity [1]n[1]_n by thm-integers-modulo-n-basic-algebra. A class uZ/nu\in\mathbb Z/n is a unit when it is invertible in that monoid (def-invertible-element). The set of all units is (Z/n)×:={uZ/n:some vZ/n satisfies uv=[1]n}.(\mathbb Z/n)^\times:=\{\,u\in\mathbb Z/n:\text{some }v\in\mathbb Z/n\text{ satisfies }uv=[1]_n\,\}. By lem-monoid-units-form-a-group, multiplication restricts to a group operation on (Z/n)×(\mathbb Z/n)^\times, called the unit group modulo nn. The quotient Z/n\mathbb Z/n is finite with cardinality nn by thm-standard-representatives-modulo-n, and its unit set is a finite subset by thm-subset-of-a-finite-set. Euler's totient function is therefore defined for every positive integer nn by φ(n):=(Z/n)×N\varphi(n):=\big|(\mathbb Z/n)^\times\big|\in\mathbb N (def-finite-cardinality). For n=1n=1, the quotient has one element, which is its multiplicative identity and hence a unit, so φ(1)=1\varphi(1)=1 follows from the definition. (The unit group (Z/n)×(\mathbb{Z}/n)^\times and Euler's totient φ(n)=(Z/n)×\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert for n1n\ge1).

[L2]

Let GG be a group and let N0,,Nr1N_0,\ldots,N_{r-1} be normal subgroups, where rNr\in\mathbb N. They form an internal direct product when they generate GG and, for each i<ri<r, NiNj:j<r, ji={e}.N_i\cap\langle N_j:j<r,\ j\ne i\rangle=\{e\}. The empty family is an internal direct product of the trivial group. For two subgroups of an abelian group this says G=HKG=HK and HK={e}H\cap K=\{e\}; in additive notation one writes G=HKG=H\oplus K. Normal subgroups and generated subgroups are those of def-normal-subgroup and def-generated-subgroup, and the comparison product is def-external-direct-product-of-groups. (Internal direct products of finitely many normal subgroups).

[L3]

Let N0,,Nr1GN_0,\ldots,N_{r-1}\trianglelefteq G. The following are equivalent: the NiN_i form an internal direct product of GG; every gGg\in G has a unique expression g=n0nr1g=n_0\cdots n_{r-1} with niNin_i\in N_i; and the multiplication map μ:i<rNiG\mu:\prod_{i<r}N_i\to G is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

[L4]

For every finite abelian group GG there is a unique list 1<n1nr1<n_1\mid\cdots\mid n_r such that GCn1××CnrG\cong C_{n_1}\times\cdots\times C_{n_r}. Moreover G=n1nr|G|=n_1\cdots n_r. The trivial group corresponds to the empty list and empty product. (Fundamental theorem of finite abelian groups: invariant-factor form).

Verification

technique · direct
1.1

Successive powers of 33 modulo 100100 are 3,9,27,81,43,29,87,61,83,49,47,41,23,69,7,21,63,89,67,1.3,9,27,81,43,29,87,61,83,49,47,41,23,69,7,21,63,89,67,1. Thus the first positive exponent giving 11 is 2020, so ord(3)=20\operatorname{ord}(3)=20.

givenL1
2.1

The class of 1-1, represented by 9999, has order 22. The list in step 1.1 contains all 2020 elements of 3\langle3\rangle and does not contain 9999, so 13-1\notin\langle3\rangle. Hence the two cyclic subgroups intersect trivially.

step 1.1
3.1

Trivial intersection makes the 202=4020\cdot2=40 products distinct. A unit representative modulo 100100 is divisible by neither 22 nor 55, since a multiple of either prime cannot have a product congruent to 11 modulo 100100. Among 0,,990,\ldots,99, inclusion-exclusion leaves 1005020+10=40100-50-20+10=40 representatives divisible by neither. Thus U(100)U(100) has at most 4040 elements, so the displayed products exhaust it. The two subgroups therefore form an internal direct product; recognition gives the isomorphism, and 2202\mid20 gives the invariant-factor order.

step 2.1L1L2L3L4

Depends on

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Sources