From the Hurwitz action to the binary tetrahedral group [fgap-000E]
✍️sourceAGENTDRAFTED
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From the Hurwitz action to the binary tetrahedral group [fgap-000E]
✍️sourceAGENTDRAFTED
- July 28, 2026
The action calculated in the Hurwitz action on the quaternion group now lets us combine the quaternion group \(Q\) and the cyclic group \(C\). We first recall the general steps: normalization makes a set product into a subgroup, complementarity gives unique factorization, and unique factorization produces an internal semidirect product.
Applying these steps inside \(\mathbb {H}^{\times }\) gives a concrete group with 24 elements: \[ Q,\ C \longrightarrow T=Q\vee C=QC \longrightarrow T\cong Q\rtimes C. \] The last isomorphism compares two models; it does not identify their elements.
Definition 1. normalizer of a subgroup [fgap-000F]AGENTDRAFTED
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Definition 1. normalizer of a subgroup [fgap-000F]AGENTDRAFTED
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Let \(N\) be a subgroup of \(G\). The normalizer of \(N\) in \(G\) is \[ N_G(N)=\{g\in G:gNg^{-1}=N\}. \] It is the largest subgroup of \(G\) in which \(N\) is normal. In particular, a subgroup \(H\leq G\) acts on \(N\) by conjugation precisely when \[ H\leq N_G(N). \] See [fre2023discrete, sec. 4.2.6, pp. 56--57].
The ambient group matters. If \(H\) normalizes \(N\), then \(N\) is normal in the subgroup generated by \(N\) and \(H\). This does not imply that \(N\) is normal in all of \(G\).
For the quaternion subgroups in Quaternion and order-three subgroups, the three conjugation calculations in the Hurwitz action on the quaternion group give \[ C\leq N_{\mathbb {H}^{\times }}(Q). \] They do not claim that \(Q\) is normal in \(\mathbb {H}^{\times }\).
Lemma 2. a normalized product is the generated subgroup [fgap-000G]AGENTDRAFTED
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Lemma 2. a normalized product is the generated subgroup [fgap-000G]AGENTDRAFTED
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Let \(N,H\leq G\). If \(H\leq N_G(N)\), then the set product \[ NH=\{nh:n\in N,\ h\in H\} \] is a subgroup and \[ N\vee H=NH. \] Moreover, \(N\) is normal in \(N\vee H\).
Proof.
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Proof.
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Normalization gives the calculation needed for closure: \[ (n_1h_1)(n_2h_2) =n_1(h_1n_2h_1^{-1})h_1h_2\in NH. \] It also gives \((nh)^{-1}=(h^{-1}n^{-1}h)h^{-1}\in NH\). Thus \(NH\) is a subgroup containing both \(N\) and \(H\), while every subgroup containing \(N\) and \(H\) contains every product \(nh\). This proves \(N\vee H=NH\). Finally, conjugation by generators from \(N\) and \(H\) preserves \(N\), so \(N\) is normal in their generated subgroup.
This is the small bridge between an action by conjugation and an internal product. Compare the normalizer and internal-product discussion in [fre2023discrete, secs. 4.2.6 and 4.2.13, pp. 56--57 and 62--63].
Definition 3. complementary subgroups [fgap-000H]AGENTDRAFTED
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Definition 3. complementary subgroups [fgap-000H]AGENTDRAFTED
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Subgroups \(N,H\leq G\) are complementary, in this ordered sense, when multiplication \[ N\times H\longrightarrow G,\qquad (n,h)\longmapsto nh \] is a bijection. Equivalently, every \(g\in G\) has a unique factorization \(g=nh\).
For subgroups, the two conditions \[ N\cap H=\{1\},\qquad NH=G \] give complementarity. Existence follows from \(NH=G\). For uniqueness, if \(n_1h_1=n_2h_2\), then \[ n_2^{-1}n_1=h_2h_1^{-1}\in N\cap H, \] so both sides are \(1\).
Trivial intersection alone is not enough: a complement must also provide the factorization of the ambient group. Fré uses these ingredients for an internal semidirect product in [fre2023discrete, sec. 4.2.13, pp. 62--63].
Theorem 4. the internal semidirect-product theorem [fgap-000I]AGENTDRAFTED
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Theorem 4. the internal semidirect-product theorem [fgap-000I]AGENTDRAFTED
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Let \(N,H\leq G\). Suppose \(N\) is normal in \(G\) and \(N,H\) are complementary. Conjugation defines an action \[ \alpha :H\longrightarrow \operatorname {Aut}(N),\qquad \alpha (h)(n)=hnh^{-1}. \] Give \(N\times H\) the multiplication \[ (n_1,h_1)(n_2,h_2) =\bigl (n_1\alpha (h_1)(n_2),h_1h_2\bigr ). \] The resulting external semidirect product is denoted \(N\rtimes _{\alpha }H\). Multiplication induces an isomorphism \[ N\rtimes _{\alpha }H\xrightarrow {\ \cong \ }G, \qquad (n,h)\longmapsto nh. \]
Proof.
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Proof.
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The displayed multiplication is chosen so that \[ (n_1h_1)(n_2h_2) =n_1(h_1n_2h_1^{-1})h_1h_2. \] Thus the multiplication map preserves products. Complementarity makes it bijective, so it is an isomorphism.
Conversely, the canonical copies of \(N\) and \(H\) in \(N\rtimes _{\alpha }H\) are complementary, and the copy of \(N\) is normal. The external multiplication and its split exact sequence are developed in [isaev2018theory, sec. 1.4.2, pp. 58--61]; compare [fre2023discrete, sec. 4.2.13, pp. 62--63]. An arbitrary extension by a normal subgroup need not split: the complement is extra structure.
Example 5. the binary tetrahedral subgroup of the quaternions [fgap-000J]AGENTDRAFTED
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Example 5. the binary tetrahedral subgroup of the quaternions [fgap-000J]AGENTDRAFTED
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Inside \(\mathbb {H}^{\times }\), let \[ T=Q\vee C. \] The Hurwitz action in the Hurwitz action on the quaternion group says that \(C\) normalizes \(Q\). Hence a normalized product is the generated subgroup gives \[ T=QC,\qquad Q\mathrel {\trianglelefteq }T. \]
The groups \(Q\) and \(C\) have orders 8 and 3. The order of their intersection divides both, so \[ Q\cap C=\{1\}. \] They are therefore complementary in \(T\). Every element has a unique normal form \[ q\omega ^r,\qquad q\in Q,\quad 0\leq r<3, \] and the internal semidirect-product theorem gives \[ Q\rtimes _{\alpha }C\cong T,\qquad \alpha (c)(q)=cqc^{-1}. \]
The action orientation is fixed by the quaternion calculation: \[ i\longmapsto k\longmapsto j\longmapsto i. \] Voight identifies this semidirect product with the Hurwitz unit group and the binary tetrahedral group in [voight2021quaternion, sec. 11.2.4, p. 168].
Example 6. the twenty-four Hurwitz units [fgap-000K]AGENTDRAFTED
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Example 6. the twenty-four Hurwitz units [fgap-000K]AGENTDRAFTED
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The unique normal forms divide \(T\) into three disjoint rows: \[ \begin {array}{c|c|c} Q&Q\omega &Q\omega ^2\\ \hline 8\text { elements}&8\text { elements}&8\text { elements}. \end {array} \] Therefore \(|T|=8\cdot 3=24\).
The first row is \[ Q=\{\pm 1,\pm i,\pm j,\pm k\}. \] Since \[ \omega =\frac {-1+i+j+k}{2},\qquad \omega ^2=\frac {-1-i-j-k}{2}, \] left multiplication by the 8 elements of \(Q\) gives \[ Q\omega \sqcup Q\omega ^2 = \left \{ \frac {\epsilon _0+\epsilon _1i+\epsilon _2j+\epsilon _3k}{2}: \epsilon _r\in \{\pm 1\} \right \}. \] The coordinates are distinct, so these are the 16 half-integral units. Together with \(Q\), they are exactly the 24 Hurwitz units listed in [voight2021quaternion, sec. 11.2, p. 166]. Thus the subgroup \(T\) is the unit group of the Hurwitz order, not merely another group of the same cardinality.
Remark 7. internal, external, and abstract models [fgap-000L]AGENTDRAFTED
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Remark 7. internal, external, and abstract models [fgap-000L]AGENTDRAFTED
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Three related objects have appeared. The internal model \(T\leq \mathbb {H}^{\times }\) has quaternions as its elements. The external semidirect product \(Q\rtimes _{\alpha }C\) has pairs \((q,c)\) as its elements. An abstract binary tetrahedral group can instead be specified without quaternion coordinates.
The isomorphism \((q,c)\mapsto qc\) compares the first two models. It is not an equality of their elements.
The next note transports the distinguished central involution and
cardinality to an abstract binary tetrahedral model. Transporting the full
factorization to factors such as QuaternionGroup 2 and
Multiplicative (ZMod 3) also needs explicit isomorphisms and
equivariance proofs. None of this follows from the order computation
\(|T|=24\) alone.
Definition 8. the distinguished involution of abstract 2T [fgap-0014]AGENTDRAFTED
Definition 8. the distinguished involution of abstract 2T [fgap-0014]AGENTDRAFTED
Let \(T\leq \mathbb {H}^{\times }\) be the binary tetrahedral group of Hurwitz units from the binary tetrahedral subgroup of the quaternions, and let \(B\) be the abstract semidirect-product model obtained from the same Hurwitz action. Choose the group isomorphism \[ \rho :B\mathbin {\cong }T \] that compares these two models. Voight identifies \(T\) with the Hurwitz unit group and with \(Q_8\rtimes \mathbb {Z}/3\mathbb {Z}\) in [voight2021quaternion, sec. 11.2.4, p. 168].
The quaternion \(-1\) belongs to \(T\); it is central, its square is 1, and it is not 1. Define the distinguished central involution of \(B\) by \[ z=\rho ^{-1}(-1). \] Since an isomorphism preserves multiplication and reflects equality, \[ z\in Z(B),\qquad z^2=1,\qquad z\neq 1. \] Thus the abstract group retains the central involution supplied by its quaternionic realization. The element \(z\) is not identified with the quaternion \(-1\); the isomorphism \(\rho \) relates them.
The same isomorphism transports cardinality. Since the calculation in
the twenty-four Hurwitz units gives \(|T|=24\),
\[
|B|=|T|=24.
\]
The corresponding Lean theorem uses Mathlib's Nat.card; its
statement is Nat.card BinaryTetrahedral.Abstract = 24.
The occurrence of \(-1\) among the 24 Hurwitz units is explicit in [voight2021quaternion, sec. 11.2, p. 166].