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Cyclic Groups and Direct Products: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-03Open item page →

The Klein four-group as the direct product of two groups of order 22

Example

The Klein four-group is

V4=(Z/2,+)×(Z/2,+).V_4=(\mathbb Z/2,+)\times(\mathbb Z/2,+).

It has the four elements (0,0),(1,0),(0,1),(1,1)(0,0),(1,0),(0,1),(1,1). Every nonidentity element has order 22, so V4V_4 is not cyclic.

Facts & Assumptions

Given: The residue-class additive group Z/2\mathbb Z/2.

[L1]

Division with remainder is available in the integers, and congruence classes modulo 22 are the quotient group with its stated addition and identity class 00 (Division with remainder in Z\mathbb{Z}: for a∈Za \in \mathbb{Z} and b>0b > 0 there are unique q,r∈Zq, r \in \mathbb{Z} with a=qb+ra = qb + r and 0≤r<b0 \le r < b, For every n∈Nn\in\mathbb N, the congruence-class group (Z/n,+)(\mathbb Z/n,+) is the quotient group (Z,+)/nZ(\mathbb Z,+)/n\mathbb Z, For every natural nn, (Z/n,+)(\mathbb{Z}/n,+) is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, Addition and multiplication on Z/n\mathbb{Z}/n by [a]n+[b]n=[a+b]n[a]_n+[b]_n=[a+b]_n and [a]n[b]n=[ab]n[a]_n[b]_n=[ab]_n).

[L2]

The external direct product of two groups is a group with componentwise operation (G×HG\times H is a group with identity (eG,eH)(e_G,e_H), coordinatewise inverses, and homomorphic coordinate projections).

[L3]

If an element has finite order, then the cyclic subgroup it generates has that many elements (If ord⁡(g)=n\operatorname{ord}(g) = n then gk=eg^{k} = e iff kk is an integer multiple of nn, the powers g0,…,gn−1g^{0}, \dots, g^{n-1} are distinct, and ⟨g⟩\langle g \rangle has exactly nn elements; if gg has infinite order then gj=gkg^{j} = g^{k} only for j=kj = k).

Verification

technique · direct
1.1

Dividing any integer aa by 22 gives a=2q+ra=2q+r with 0≤r<20\le r<2, so its class is either [0][0] or [1][1]. These are distinct because 1−01-0 is not a multiple of 22, and [1]+[1]=[0][1]+[1]=[0]. Hence the four displayed pairs are exactly the elements of V4V_4.

L1givenalgebra
2.1

For each nonzero pair (a,b)(a,b) in that list, (a,b)+(a,b)=(a+a,b+b)=(0,0)(a,b)+(a,b)=(a+a,b+b)=(0,0); it is not the identity, so its order is 22.

step 1.1L2algebra
3.1

If V4V_4 were cyclic, a generator could not be the identity, so step 2.1 would give it order 22. Then [L3] says that its cyclic subgroup has two elements, contradicting step 1.1. Thus V4V_4 is not cyclic.

step 1.1step 2.1L3∎
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-03Open item page →

The cyclic group Z/4\mathbb Z/4 is not isomorphic to Z/2×Z/2\mathbb Z/2\times\mathbb Z/2

Counterexample

The groups Z/4\mathbb Z/4 and Z/2×Z/2\mathbb Z/2\times\mathbb Z/2 both have four elements, but they are not isomorphic: the first has an element of order 44, whereas every nonidentity element of the second has order 22.

Facts & Assumptions

Given: The additive quotient groups Z/4\mathbb Z/4 and Z/2\mathbb Z/2.

[L1]

The residue classes modulo nn form the quotient group of the additive integers (For every n∈Nn\in\mathbb N, the congruence-class group (Z/n,+)(\mathbb Z/n,+) is the quotient group (Z,+)/nZ(\mathbb Z,+)/n\mathbb Z).

[L2]

In a direct product, a pair with finite component orders m,nm,n has order the unique positive natural whose canonical integer image is lcm⁡(ι(m),ι(n))\operatorname{lcm}(\iota(m),\iota(n)) (If gg and hh have finite orders mm and nn, then ι(ord⁡(g,h))=lcm⁡(ι(m),ι(n))\iota(\operatorname{ord}(g,h))=\operatorname{lcm}(\iota(m),\iota(n)) in G×HG\times H).

[L3]

A group isomorphism is a bijective homomorphism, and a homomorphism preserves powers (Group isomorphisms, automorphisms and the set Aut⁡(G)\operatorname{Aut}(G), A group homomorphism automatically satisfies f(e)=e′f(e) = e' and f(g−1)=f(g)−1f(g^{-1}) = f(g)^{-1}, and f(gn)=f(g)nf(g^{n}) = f(g)^{n} for every n∈Zn \in \mathbb{Z}; for monoid homomorphisms preservation of the identity must be assumed).

Refutation

technique · direct
1.1

The class [1][1] in Z/4\mathbb Z/4 has order 44, while the nonzero class in Z/2\mathbb Z/2 has order 22.

L1givenalgebra
2.1

By [L2], each nonidentity pair has component orders 1,21,2 or 2,22,2. Their least common multiple is 22, so every nonidentity pair has order 22.

step 1.1L2given
3.1

A group isomorphism preserves the least positive exponent at which a power is the identity, by [L3]; it therefore cannot send the order-44 element of step 1.1 to any element of the target.

step 1.1step 2.1L3
4.1

Hence Z/4\mathbb Z/4 and Z/2×Z/2\mathbb Z/2\times\mathbb Z/2 are not isomorphic.

step 3.1∎

Sources

Standard references

Recommended treatments; not extraction sources.