Semidirect Products, Automorphism Groups and Split Extensions — Examples

The subgroup $N=⟨r⟩=\{1,(123),(132)\}$ has index two in the six-element group $S_3$, and direct conjugation by every permutation preserves the set of the two $3$-cycles. Hence $N⊴S_3$.

The subgroup $H=⟨s⟩$ has order two, intersects $N$ trivially, and the six products $r^i{s}^j$ are distinct. Thus $S_3=NH$.

Since $sr{s}^{-1}=(132)=r^{-1}$, [L1] gives the asserted semidirect product, which is the order-six case of [L2].

Each nonzero $a$ acts on $(\mathbb{R},+)$ by the automorphism $x\mapsto ax$, and multiplication of scalars composes these automorphisms. Thus [L1] gives the group law $(b,a)(d,c)=(b+ad,ac)$.

Associate $(b,a)$ with $f_{b,a}(x)=ax+b$. Composition satisfies $f_{b,a}\circ f_{d,c}(x)=acx+(ad+b)$, so the association is a homomorphism by step 1.1. It is bijective because an affine map uniquely determines its slope $a$ and intercept $b$. This is the affine action described in [L2].

By [L1]--[L3], $A_n$ is normal. The transposition $\tau$ has sign $-1$, so $⟨\tau ⟩=\{1,\tau \}$ intersects $A_n$ trivially.

If $\sigma \in S_n$ is even, then $\sigma \in A_n$. If it is odd, then $\sigma \tau$ is even and $\sigma =(\sigma \tau )\tau$. Hence $S_n=A_n⟨\tau ⟩$.

The recognition theorem [L4] gives the asserted decomposition, with the action on $A_n$ given by conjugation by $\tau$.

Every $a\in C_2\times C_2$ satisfies $a^2=1$, hence $a^{-1}=a$. The inversion automorphism is therefore the identity.

The action in [L1] is trivial by step 1.1, so [L2] gives the displayed direct product.

By [L3], the four units modulo $8$ are $1,3,5,7$. Thus [L1] and [L2] identify every holomorph element with one of the displayed affine maps.

If $ax+b=cx+d$ for every $x$, evaluation at $0$ gives $b=d$, and evaluation at $1$ then gives $a=c$. Hence the $8\cdot 4=32$ maps are distinct.

Their composition is $(x\mapsto ax+b)\circ (x\mapsto cx+d)=x\mapsto acx+(ad+b)$, which agrees with the holomorph multiplication from [L1].

By [L2], $(\mathbb{Z}\mathrm{/}8{)}^{\times }=\{1,3,5,7\}$. Each nonidentity element squares to $1$ modulo $8$: $3^2\equiv 5^2\equiv 7^2\equiv 1$.

The map from $C_2\times C_2$ sending $(0,0),(1,0),(0,1),(1,1)$ to $1,3,5,7$, respectively, is bijective, and direct multiplication modulo $8$ verifies that it preserves the group operation. Thus $(\mathbb{Z}\mathrm{/}8{)}^{\times }\cong C_2\times C_2$, and [L1] gives the result.

Integer inverses are $A^{-1}=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$ and $B^{-1}=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Thus both arrays define automorphisms by [L1].

Direct multiplication gives $AB=\left(\begin{array}{cc}1& -1\\ 1& 0\end{array}\right)$ and $BA=\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$. They differ, so the corresponding automorphisms do not commute.

Every automorphism of $V$ fixes the identity and permutes the three nonidentity elements. Conversely, any permutation of those three elements preserves the group law: the product of two distinct nonidentity elements is the third. Thus $\mathrm{Aut}(V)\cong S_3$ and has order six.

By [L1], $\mathrm{\mid }\mathrm{Hol}(V)\mathrm{\mid }=4\cdot 6=24$. By [L2] and [L3], its faithful action embeds it into $S_4$, which also has $4!=24$ elements by [L4].

An injective map between these two finite sets of equal size is surjective. Hence the embedding is an isomorphism.

Since $3\mid 6$, [L1] gives a unique nonabelian isomorphism type of order $21$.

Modulo $7$, one has $2^3\equiv 1$ while $2\equiv ̸1$, so $[2]$ has order three in $(\mathbb{Z}\mathrm{/}7{)}^{\times }$. By [L2], it defines a nontrivial action of $C_3$ on $C_7$.

The resulting semidirect product is therefore the unique nonabelian group from [L1].

The injection sends $[a{]}_p$ to $[pa{]}_{p^2}$, and the quotient map $\pi$ sends $[x{]}_{p^2}$ to $[x{]}_p$. The kernel of $\pi$ is the subgroup $p{C}_{p^2}$ of order $p$, so the sequence is short exact.

Suppose, for contradiction, that a section $s:C_p\to C_{p^2}$ existed. Since $\pi s$ is the identity, $s$ is injective, so its image $H$ is a subgroup of order $p$. By [L2], $H=⟨[a{]}_{p^2}⟩$ for some $a$. Its generator has order $p$, so [L3] gives $p^2\mid pa$ and hence $p\mid a$. Thus $H\subseteq p{C}_{p^2}$; both subgroups have $p$ elements, so they are equal.

Then $\pi s$ is the zero homomorphism, contradicting that it is the identity on the nontrivial group $C_p$. Thus the extension does not split.

$⟨i⟩⊴Q_8$ and $\mathrm{\mid }{Q}_8\mathrm{/}⟨i⟩\mathrm{\mid }=2$. By [L1] and [L5], $[Q_8:⟨i⟩]=8\mathrm{/}4=2$, so [L3] makes $⟨i⟩$ normal and [L4] gives $\mathrm{\mid }{Q}_8\mathrm{/}⟨i⟩\mathrm{\mid }=2$.

The middle group $Q_8$ is nonabelian: $ij=k$ and $ji=-k$ with $k\mathrm{\ne }-k$ by [L8], and $i,j,k,-k$ all lie in $Q_8$ by [L1]. The middle group of the cyclic witness $1\to C_p\to C_{p^2}\to C_p\to 1$ has order $p^2$ and is therefore abelian by [L10], so the refutation below is not a restatement of that one.

The projection $\pi :Q_8\to Q_8\mathrm{/}⟨i⟩$, $\pi (g)=g⟨i⟩$, is a surjective homomorphism with $\ker \pi =⟨i⟩$. It is defined because $⟨i⟩$ is normal by step 1.1, it is a homomorphism because [L9] gives $(g⟨i⟩)(h⟨i⟩)=gh⟨i⟩$, it is surjective because every element of $Q_8\mathrm{/}⟨i⟩$ is a coset $g⟨i⟩$ by [L9], and its kernel is $\{g:g⟨i⟩=⟨i⟩\}=⟨i⟩$, the identity of the quotient being $⟨i⟩$ itself by [L9].

Suppose, for contradiction, that a section $s:Q_8\mathrm{/}⟨i⟩\to Q_8$ existed. Since $\pi \circ s=\mathrm{id}$, the map $s$ is injective, so its image $K=s(Q_8\mathrm{/}⟨i⟩)$ is a subgroup of $Q_8$ by [L7], of the same size as $Q_8\mathrm{/}⟨i⟩$, hence $\mathrm{\mid }K\mathrm{\mid }=2$ by step 1.1. Writing $K=\{1,x\}$ with $x\mathrm{\ne }1$, [L11] applied to the finite group $K$ gives that $\mathrm{ord}(x)$ divides $\mathrm{\mid }K\mathrm{\mid }=2$, so $\mathrm{ord}(x)$ is $1$ or $2$; it is not $1$, since by [L7] that would make $x=x^1=1$. Hence $\mathrm{ord}(x)=2$.

The sequence is short exact. Write $\phi :⟨i⟩\to Q_8$ for the inclusion; it is an injective homomorphism with $\mathrm{im}\phi =⟨i⟩$, and step 2.1 gives $\ker \pi =⟨i⟩$, so $\mathrm{im}\phi =\ker \pi$ as [L2] requires. The kernel is $⟨i⟩$, cyclic of order $4$ because it is generated by $i$; the quotient has order $2$ by step 1.1, and $2$ is prime, so [L6] makes it cyclic.

By [L1] the only element of $Q_8$ of order $2$ is $-1$, so $x=-1$. But $-1\in ⟨i⟩=\ker \pi$ by [L1] and step 2.1, hence $\pi (x)$ is the identity of $Q_8\mathrm{/}⟨i⟩$.

On the other hand $x=s(h)$ for the unique non-identity $h\in Q_8\mathrm{/}⟨i⟩$, because $s$ is injective and $s$ sends the identity to the identity by [L7]; and then $\pi (x)=\pi (s(h))=h\mathrm{\ne }1$, contradicting step 3.2. Hence no section exists and the extension does not split. [step 2.2, step 3.2, L2, L7, discharge-contradiction] $\mathrm{■}$

The subgroup $H=⟨s⟩$ is the canonical complement to $⟨r⟩$. The inversion action on $C_3$ is nontrivial because $r^{-1}=r^2\mathrm{\ne }r$.

Therefore $H$ is not normal by [L2]. Explicitly, $rs{r}^{-1}=r^{2}s$, which is not in $\{1,s\}$ by the uniqueness in [L1].

The coordinate swap $u(x,y)=(y,x)$ preserves addition and is its own inverse, so it is an automorphism by [L2].

But $u(K)=⟨(0,1)⟩\mathrm{\ne }K$. Hence $K$ is not characteristic by [L1], even though $V$ is abelian and therefore every subgroup of $V$ is normal.

The kernel of reduction $\pi :C_{p^2}\to C_p$ is $p{C}_{p^2}$, which has order $p$ by [L3], so the displayed sequence is short exact.

Assume, for contradiction, that [A1] holds and let $s$ be a section. Since $\pi s$ is the identity, $s$ is injective and its image $H$ has order $p$. By [L2], $H=⟨[a{]}_{p^2}⟩$ for some $a$. Its generator has order $p$, so [L3] gives $p^2\mid pa$ and hence $p\mid a$. Thus $H\subseteq p{C}_{p^2}$; both subgroups have $p$ elements, so $H=\ker \pi$.

Hence $\pi s$ is zero, contradicting [L1]. Therefore this sequence does not split and [A1] is false.

The cyclic group $C_{2p}$ contains its index-two subgroup $C_p$, and quotienting by it gives $C_2$. Thus it is the middle group of an extension of $C_2$ by $C_p$ in the sense of [L1].

By [L2], the rotation subgroup $C_p$ is normal in $D_p$ and the complementary reflection subgroup maps isomorphically to the quotient $C_2$. Hence $D_p$ is another middle group for the same kernel and quotient.

The group $C_{2p}$ is abelian, while $D_p$ is not: inversion on $C_p$ is nontrivial because $p$ is odd. Therefore they are not isomorphic, refuting [A1].

Every automorphism of $V$ fixes the identity and permutes the three nonidentity elements. Conversely, every permutation of those elements preserves the group law because the sum of two distinct nonidentity elements is the third. Thus restriction gives $\mathrm{Aut}(V)\cong S_3$ by [L1] and [L2].

In $S_3$, the transpositions $(12)$ and $(23)$ do not commute. Hence $\mathrm{Aut}(V)$ is nonabelian, contradicting [A1] and refuting the claim.