PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$Q_{8}$ is a subgroup of $H^{\times}$ with eight elements, and $- 1$ is its only element of order $2$

Statement

Let $Q_{8} = {1, - 1, i, - i, j, - j, k, - k} \subseteq H^{\times}$ be as in The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions. Then:

$Q_{8}$ is a subgroup of $H^{\times}$ and $∣ Q_{8} ∣ = 8$ ;
$1$ is the only element of order $1$ , $- 1$ is the only element of order $2$ , and each of $\pm i, \pm j, \pm k$ has order $4$ ;
$⟨ i ⟩ = {1, i, - 1, - i}$ is a subgroup of order $4$ containing $- 1$ , and the same holds for $⟨ j ⟩$ and $⟨ k ⟩$ .

Facts & Assumptions

Given: The quaternions $H$ , the basis quaternions $1, i, j, k$ , the real embedding $λ \mapsto \hat{λ}$ , the element $n = \hat{- 1}$ and the abbreviation $- x = n x$ of The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions and The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ .

[F1]

Evaluating the multiplication formula of $H$ on the basis quaternions gives $i^{2} = j^{2} = k^{2} = - 1$ , $i j = k$ , $j k = i$ , $k i = j$ , $j i = - k$ , $k j = - i$ , $i k = - j$ , together with $1 x = x 1 = x$ for $x \in {1, i, j, k}$ ; and $\hat{λ} x = x \hat{λ} = (λ x_{0}, λ x_{1}, λ x_{2}, λ x_{3})$ for every real $λ$ (The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ).

[F2]

$H$ is a ring with identity $1_{H}$ , and $H ∖ {0_{H}}$ is a group under multiplication ( $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ ).

[F3]

A nonempty subset $H \subseteq G$ of a group $G$ is a subgroup exactly when $g h^{- 1} \in H$ for all $g, h \in H$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup).

[F4]

The order $ord (g)$ of an element $g$ of a group is the least natural $m \geq 1$ with $g^{m} = e$ when such an $m$ exists, and $\infty$ otherwise (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[F5]

If $ord (g) = m$ with $m \geq 1$ , then for every integer $t$ one has $g^{t} = e$ if and only if $m ∣ t$ ; the powers $g^{0}, \dots, g^{m - 1}$ are pairwise distinct; and $⟨ g ⟩ = {g^{s} : s < m}$ has exactly $m$ 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 $j = k$ ).

[F6]

For $g$ in a group, $⟨ g ⟩$ is the set of integer powers of $g$ ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian, The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

Proof

technique · direct

1.1

The element $n = \hat{- 1}$ is central in $H$ and satisfies $n^{2} = 1$ . Centrality is the clause $\hat{λ} x = x \hat{λ}$ of [F1] with $λ = - 1$ ; and $n^{2} = \hat{- 1} \hat{- 1}$ , which the same clause evaluates at $λ = - 1$ and $x = (- 1, 0, 0, 0)$ to give $(1, 0, 0, 0) = 1$ . Hence also $- (- x) = n (n x) = n^{2} x = x$ for every $x$ , multiplication in $H$ being associative by [F2].

F1F2algebra

2.1

The eight quaternions listed in $Q_{8}$ are pairwise distinct, so $∣ Q_{8} ∣ = 8$ . Written out as quadruples they are $(\pm 1, 0, 0, 0)$ , $(0, \pm 1, 0, 0)$ , $(0, 0, \pm 1, 0)$ , $(0, 0, 0, \pm 1)$ , and two such quadruples agree only if they agree in every coordinate; since $1 \neq - 1$ and $1 \neq 0$ in $R$ , no two of the eight agree.

step 1.1F1algebra

2.2

$Q_{8}$ is closed under multiplication. Every element of $Q_{8}$ is $ε u$ with $u \in {1, i, j, k}$ and $ε \in {1, n}$ , and for two such elements associativity and the centrality of step 1.1 give $(ε u) (δ v) = (ε δ) (u v)$ . Here $ε δ \in {1, n}$ because $n^{2} = 1$ , and $u v \in {\pm 1, \pm i, \pm j, \pm k}$ by the table in [F1]. Hence $(ε u) (δ v) \in Q_{8}$ .

step 1.1F1F2

3.1

Every element of $Q_{8}$ has an inverse lying in $Q_{8}$ . From [F1], $i (- i) = - (i^{2}) = - (- 1) = 1$ by step 1.1, and likewise $j (- j) = k (- k) = 1$ ; also $1 \cdot 1 = 1$ and $n n = 1$ . So each of $1, - 1$ is its own inverse and each of $\pm i, \pm j, \pm k$ has its negative as inverse, all inside $Q_{8}$ .

step 1.1step 2.2F1

3.2

$ord (1) = 1$ and $ord (- 1) = 2$ . The identity of $Q_{8}$ is $1$ by [F2], so $ord (1) = 1$ by [F4]. For $- 1 = n$ we have $n \neq 1$ by step 2.1 and $n^{2} = 1$ by step 1.1, so $2$ is the least $m \geq 1$ with $n^{m} = 1$ .

step 1.1step 2.1F2F4

3.3

Each of $\pm i, \pm j, \pm k$ has order $4$ . Write such an element as $ε u$ with $u \in {i, j, k}$ and $ε \in {1, n}$ . By step 1.1 and [F1], $(ε u)^{2} = ε^{2} u^{2} = u^{2} = - 1$ , and $(ε u)^{4} = ((ε u)^{2})^{2} = (- 1)^{2} = 1$ . So the order is finite by [F4] and divides $4$ by [F5]; it is not $1$ because $(ε u)^{1} = ε u \neq 1$ by step 2.1, and not $2$ because $(ε u)^{2} = - 1 \neq 1$ by step 2.1. The only remaining divisor of $4$ is $4$ .

step 1.1step 2.1F1F4F5

4.1

$Q_{8}$ is a subgroup of $H^{\times}$ . It is nonempty and contained in $H^{\times}$ , since none of its eight elements is $0_{H}$ by step 2.1; and for $g, h \in Q_{8}$ steps 2.2 and 3.1 give $h^{- 1} \in Q_{8}$ and then $g h^{- 1} \in Q_{8}$ , which is the criterion [F3]. With step 2.1 this proves claim 1.

step 2.1step 2.2step 3.1F3

4.2

Claim 2 follows: steps 3.2 and 3.3 assign an order to each of the eight elements of $Q_{8}$ , and among them exactly one has order $1$ , namely $1$ , and exactly one has order $2$ , namely $- 1$ .

step 2.1step 3.2step 3.3

5.1

$⟨ i ⟩ = {1, i, - 1, - i}$ has four elements and contains $- 1$ . By step 3.3 $ord (i) = 4$ , so [F5] gives $⟨ i ⟩ = {i^{0}, i^{1}, i^{2}, i^{3}}$ with these four powers pairwise distinct, and [F6] confirms these are all the integer powers. Evaluating, $i^{0} = 1$ , $i^{1} = i$ , $i^{2} = - 1$ by [F1], and $i^{3} = i^{2} i = (- 1) i = - i$ by step 1.1. The same computation with $j$ and with $k$ gives $⟨ j ⟩ = {1, j, - 1, - j}$ and $⟨ k ⟩ = {1, k, - 1, - k}$ , each of order $4$ and each containing $- 1$ . This is claim 3. [step 1.1, step 3.3, F1, F5, F6] $■$

Depends on

Used by

1→⟨ i⟩→ Q₈→ Q₈/⟨ i⟩→1 does not split, with nonabelian middle group Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 97 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. S. Milne, Group Theory, Example 3.9(c) (standard reference, not scraped)

Q8Q8​ is a subgroup of H×H× with eight elements, and −1−1 is its only element of order 22