Semidirect Products, Automorphism Groups and Split Extensions

For three pairs, multiplication in either parenthesisation gives $(n{\alpha }_h(n^{\mathrm{\prime }}){\alpha }_{h{h}^{\mathrm{\prime }}}(n^{\mathrm{\prime }\mathrm{\prime }}),h{h}^{\mathrm{\prime }}{h}^{\mathrm{\prime }\mathrm{\prime }})$ because ${\alpha }_h$ is a homomorphism and ${\alpha }_{h{h}^{\mathrm{\prime }}}={\alpha }_h{\alpha }_{h^{\mathrm{\prime }}}$. Thus the operation is associative.

Since ${\alpha }_1$ is the identity and every ${\alpha }_h$ preserves $1_N$, the pair $(1_N,1_H)$ is a two-sided identity.

Put $y={\alpha }_{h^{-1}}(n^{-1})$. Then $(n,h)(y,h^{-1})=(n{\alpha }_h(y),1_H)=(1_N,1_H)$ and $(y,h^{-1})(n,h)=(y{\alpha }_{h^{-1}}(n),1_H)=(1_N,1_H)$. Hence the displayed pair is the two-sided inverse.

The multiplication and inverse formulas in [L1] show that both displayed sets contain the identity and are closed under products and inverses. Hence both are subgroups by [L2].

Direct multiplication gives $(n,1)(1,h)=(n,h)$, while equality $(n,1)=(1,h)$ forces $n=1$ and $h=1$. This proves existence and uniqueness of the factorisation and the trivial intersection.

Using the inverse formula gives $(1,h)(n,1)(1,h{)}^{-1}=({\alpha }_h(n),1)$. Conjugation by an element of $\bar{N}$ also preserves $\bar{N}$ because it is a subgroup. By step 1.2 every group element is a product of an element of $\bar{N}$ and one of $\bar{H}$, so its conjugation preserves $\bar{N}$; applying the same argument to its inverse gives equality. Thus $\bar{N}$ is normal by [L2], and the displayed calculation identifies the induced action with $\alpha$.

[reverse] Suppose the action is trivial. The semidirect law becomes $(n,h)(n^{\mathrm{\prime }},h^{\mathrm{\prime }})=(n{n}^{\mathrm{\prime }},h{h}^{\mathrm{\prime }})$, which is the direct-product law from [L2].

[forward] Suppose the canonical decomposition is an internal direct product, so $\bar{H}$ as well as $\bar{N}$ is normal. For $x\in \bar{N}$ and $y\in \bar{H}$, normality gives $xy{x}^{-1}{y}^{-1}\in \bar{H}$ and also $xy{x}^{-1}{y}^{-1}=x(y{x}^{-1}{y}^{-1})\in \bar{N}$. Thus this commutator lies in $\bar{N}\cap \bar{H}=\{1\}$ by [L1], so $x$ and $y$ commute.

The conjugation formula in [L1] now gives $({\alpha }_h(n),1)=(n,1)$ for every $n,h$. Hence every ${\alpha }_h$ is the identity.

[forward] Assume the three conditions in [L1]. By [L2], conjugation defines an action $\alpha :H\to \mathrm{Aut}(N)$, so the domain of $\mathrm{\Phi }$ is a group by [L3].

For $(n,h),(n^{\mathrm{\prime }},h^{\mathrm{\prime }})$ one has

If $nh=n^{\mathrm{\prime }}{h}^{\mathrm{\prime }}$, then $n^{-1}{n}^{\mathrm{\prime }}=h{h}^{\mathrm{\prime }-1}\in N\cap H$, hence both sides are $1$; therefore $n=n^{\mathrm{\prime }}$ and $h=h^{\mathrm{\prime }}$. Thus $\mathrm{\Phi }$ is injective, and step 1.2 makes it bijective and therefore an isomorphism by [L5].

[reverse] Conversely, assume conjugation restricts to $\alpha :H\to \mathrm{Aut}(N)$ and that $\mathrm{\Phi }$ is such an isomorphism; the first assumption is what makes $N{⋊}_{\alpha }H$, and hence $\mathrm{\Phi }$, defined. Transport the canonical-factor properties from [L4] through $\mathrm{\Phi }$. The images are $N,H$, so the three conditions in [L1] hold.

Suppose $s$ is a section and put $K=s(H)$. If $s(h)\in \ker \pi$, then $h=\pi s(h)=1$, so $s$ is injective by [L4] and $K\cap \ker \pi =\{1\}$.

Conversely, suppose $K$ is a complement. The restriction $\pi {\mathrm{\mid }}_K$ is injective because its kernel is $K\cap \ker \pi$, and it is surjective because $G=(\ker \pi )K$ and $\pi$ kills the first factor. Thus it is an isomorphism by [L3] and [L4].

For $g\in G$, put $h=\pi (g)$. Then $gs(h{)}^{-1}\in \ker \pi$, so $g\in (\ker \pi )K$. Hence $K$ is a complement, and [L2] gives the compatible semidirect-product decomposition with the stated conjugation action.

The inverse $s=(\pi {\mathrm{\mid }}_K{)}^{-1}:H\to K\hookrightarrow G$ is a homomorphic section. Finally, any compatible external semidirect decomposition supplies its canonical complement and hence a section by the same construction.

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

The eight quaternions listed in $Q_8$ are pairwise distinct, so $\mathrm{\mid }{Q}_8\mathrm{\mid }=8$. 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 $1\mathrm{\ne }-1$ and $1\mathrm{\ne }0$ in $\mathbb{R}$, no two of the eight agree.

$Q_8$ is closed under multiplication. Every element of $Q_8$ is $\epsilon u$ with $u\in \{1,i,j,k\}$ and $\epsilon \in \{1,n\}$, and for two such elements associativity and the centrality of step 1.1 give $(\epsilon u)(\delta v)=(\epsilon \delta )(uv)$. Here $\epsilon \delta \in \{1,n\}$ because $n^2=1$, and $uv\in \{\pm 1,\pm i,\pm j,\pm k\}$ by the table in [F1]. Hence $(\epsilon u)(\delta v)\in Q_8$.

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 $nn=1$. 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 $Q_8$.

$\mathrm{ord}(1)=1$ and $\mathrm{ord}(-1)=2$. The identity of $Q_8$ is $1$ by [F2], so $\mathrm{ord}(1)=1$ by [F4]. For $-1=n$ we have $n\mathrm{\ne }1$ by step 2.1 and $n^2=1$ by step 1.1, so $2$ is the least $m\ge 1$ with $n^m=1$.

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

$Q_8$ is a subgroup of ${\mathbb{H}}^{\times }$. It is nonempty and contained in ${\mathbb{H}}^{\times }$, since none of its eight elements is $0_{\mathbb{H}}$ 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.

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 $-1$.

$⟨i⟩=\{1,i,-1,-i\}$ has four elements and contains $-1$. By step 3.3 $\mathrm{ord}(i)=4$, 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] $\mathrm{■}$

Fix $g\in G$. Normality of $N$ and [L2] show that conjugation by $g$ maps $N$ to itself; by [L3], its restriction is an automorphism of $N$.

Since $K$ is characteristic in $N$, [L1] gives $gK{g}^{-1}=K$. This holds for every $g\in G$, so $K⊴G$ by [L2].

Each map $x\mapsto g\alpha (x)$ is a permutation, with inverse $x\mapsto {\alpha }^{-1}(g^{-1}x)$.

Composition gives $\rho (g,\alpha )(\rho (h,\beta )(x))=g\alpha (h)(\alpha \beta )(x)$, which equals $\rho ((g,\alpha )(h,\beta ))(x)$ by [L1]. Therefore $\rho$ is an action, equivalently a homomorphism to $\mathrm{Sym}(G)$, by [L2] and [L3].

If $\rho (g,\alpha )$ is the identity permutation, evaluation at $1_G$ gives $g=1_G$. Then $\alpha (x)=x$ for every $x\in G$, so $\alpha ={\mathrm{id}}_G$. Thus the action is faithful by [L2], and the corresponding homomorphism is injective: equality of two images reduces, after multiplying by an inverse, to this identity case.

Let $S=K_{{\omega }_0}$. Regularity gives, for each $k\in K$, a unique $r\in R$ with $r{\omega }_0=k{\omega }_0$. Then $r^{-1}k\in S$, so $K=RS$ and $R\cap S=\{1\}$.

For $s\in S$ and $r\in R$, normality gives $sr{s}^{-1}\in R$ and $s(r{\omega }_0)=(sr{s}^{-1}){\omega }_0$. Thus, under the chosen identification, $s$ acts as the automorphism $r\mapsto sr{s}^{-1}$, while $R$ acts by left translations.

Since $R⊴K$, [L2] identifies $K$ with $R⋊S$, where $S$ acts on $R$ by conjugation.

The resulting permutations are precisely of the affine form in [L3], giving a homomorphism $K\to \mathrm{Hol}(R)$. It is injective because the original action is faithful.

By [L6] every element of $C_n$ is a power $g^k$, and a homomorphism $f$ satisfies $f(g^k)=f(g{)}^k$, so $f$ is determined by $f(g)$; writing $f(g)=g^a$, every endomorphism has the form $f_a(g^k)=g^{ak}$. By [L3], $f_a=f_b$ exactly when $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}n)$, so the endomorphisms are indexed by the residue classes of $\mathbb{Z}\mathrm{/}n$, which [L1] identifies with $C_n$ as an additive group.