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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 $2$

Example

The Klein four-group is

$V_{4} = (Z / 2, +) \times (Z / 2, +) .$

It has the four elements $(0, 0), (1, 0), (0, 1), (1, 1)$ . Every nonidentity element has order $2$ , so $V_{4}$ is not cyclic.

Facts & Assumptions

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

[L1]

Division with remainder is available in the integers, and congruence classes modulo $2$ are the quotient group with its stated addition and identity class $0$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ , For every $n \in N$ , the congruence-class group $(Z / n, +)$ is the quotient group $(Z, +) / n Z$ , For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, Addition and multiplication on $Z / n$ by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ ).

[L2]

The external direct product of two groups is a group with componentwise operation ( $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

[L3]

If an element has finite order, then the cyclic subgroup it generates has that many elements (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Verification

technique · direct

1.1

Dividing any integer $a$ by $2$ gives $a = 2 q + r$ with $0 \leq r < 2$ , so its class is either $[0]$ or $[1]$ . These are distinct because $1 - 0$ is not a multiple of $2$ , and $[1] + [1] = [0]$ . Hence the four displayed pairs are exactly the elements of $V_{4}$ .

L1givenalgebra

2.1

For each nonzero pair $(a, b)$ in that list, $(a, b) + (a, b) = (a + a, b + b) = (0, 0)$ ; it is not the identity, so its order is $2$ .

step 1.1L2algebra

3.1

If $V_{4}$ were cyclic, a generator could not be the identity, so step 2.1 would give it order $2$ . Then [L3] says that its cyclic subgroup has two elements, contradicting step 1.1. Thus $V_{4}$ is not cyclic.

step 1.1step 2.1L3∎

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

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

Counterexample

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

Facts & Assumptions

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

[L1]

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

[L2]

In a direct product, a pair with finite component orders $m, n$ has order the unique positive natural whose canonical integer image is $lcm (ι (m), ι (n))$ (If $g$ and $h$ have finite orders $m$ and $n$ , then $ι (ord (g, h)) = lcm (ι (m), ι (n))$ in $G \times H$ ).

[L3]

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

Refutation

technique · direct

1.1

The class $[1]$ in $Z / 4$ has order $4$ , while the nonzero class in $Z / 2$ has order $2$ .

L1givenalgebra

2.1

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

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- $4$ element of step 1.1 to any element of the target.

step 1.1step 2.1L3

4.1

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

step 3.1∎

Cyclic Groups and Direct Products: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

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

Example

Facts & Assumptions

Verification

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

Counterexample

Facts & Assumptions

Refutation

Sources

The Klein four-group as the direct product of two groups of order $2$

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