Group Extensions Complements and Schur Zassenhaus — Examples

Reduction mod $2$ is a surjective homomorphism $\pi :C_4\to C_2$, and the displayed map $s$ is a set-theoretic section because $\pi (s([0]))=[0]$ and $\pi (s([1]))=[1]$.

But $s([1]+[1])=s([0])=[0]\mathrm{\ne }[2]=s([1])+s([1])$ in $C_4$. Hence $s$ is not a homomorphism. Therefore the universal claim [L1] is false.

By [L1], the rotation subgroup $⟨r⟩$ is a normal copy of $C_4$ and the reflection subgroup $⟨s⟩$ is a copy of $C_2$. Their intersection is trivial and they generate $D_4$.

Thus $⟨s⟩$ is a complement to $⟨r⟩$ in $D_4$, so [L2] gives the displayed split extension of $C_4$ by $C_2$.

Let $N=C_2\times \{1\}$ and $H=\{1\}\times C_2$. By [L2], these are subgroups of $V$ with trivial intersection and product $V$, so they define a split extension of $C_2$ by $C_2$.

Again by [L2], elements of $N$ and $H$ commute coordinatewise. Therefore the complement $H$ centralizes the kernel $N$, and [L1] makes the extension direct.

The subgroup $V$ has order $4$ and index $3$, so it is a normal Hall subgroup of $A_4$. The four subgroups $⟨(123)⟩$, $⟨(124)⟩$, $⟨(134)⟩$, and $⟨(234)⟩$ each have order $3$, intersect $V$ trivially, and together with $V$ generate $A_4$, so they are complements.

There are exactly four subgroups of order $3$ in $A_4$, because the eight $3$-cycles come in inverse pairs and each order-$3$ subgroup has exactly two nonidentity elements. Hence the four displayed complements are all the complements to $V$. Since $A_4\mathrm{/}V\cong C_3$ is solvable, [L2] also predicts that they form one conjugacy class.

The subgroup $A_3$ has order $3$ and index $2$, so each subgroup generated by a transposition intersects it trivially and together they generate $S_3$. Thus the three transposition subgroups are complements to $A_3$.

They are conjugate by direct calculation: $(123)(12)(123{)}^{-1}=(23)$ and $(132)(12)(132{)}^{-1}=(13)$. Since $S_3\mathrm{/}{A}_3\cong C_2$ is solvable, this also matches [L1].

Let ${\pi }_1,{\pi }_2:E\to Q$ be the surjective homomorphisms ${\pi }_1(\bar{x})=\bar{x}(\mathrm{m}\mathrm{o}\mathrm{d}3),{\pi }_2(\bar{x})=2\bar{x}(\mathrm{m}\mathrm{o}\mathrm{d}3)\mathrm{.}$ Both have kernel $\mathrm{im}j=\{\bar{0},\bar{3},\bar{6}\}$, so $0\to N\stackrel{j}{\to }E\stackrel{{\pi }_1}{\to }Q\to 0,0\to N\stackrel{j}{\to }E\stackrel{{\pi }_2}{\to }Q\to 0$ are two extensions of $Q$ by $N$ with the same middle group $E$.

The middle groups are literally identical, but the two extension structures are not equivalent. Indeed, any automorphism of $E=\mathbb{Z}\mathrm{/}9\mathbb{Z}$ is multiplication by a unit $u\in \{1,2,4,5,7,8\}$. If it fixed the kernel inclusion, then $u\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}3)$; if it also carried ${\pi }_1$ to ${\pi }_2$, then $2u\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}3)$, so $u\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}3)$, impossible. Thus [L1] is false.