Problem 5.5.4 ($Z_4 \rtimes_\varphi Z_2$ and $(Z_2 \times Z_2) \rtimes_\varphi Z_2$ are isomorphic to $D_8$)

Proof. We must prove the existence of the non-abelian semidirect products $\mathbb{Z}∕4\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$ and $(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}){⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$, and show that they are isomorphic to $D_8$.

• Construction of $\mathbb{Z}∕4\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$.

  There is a unique non trivial homomorphism

  $$\phi :\mathbb{Z}∕2\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕4\mathbb{Z})\simeq {(\mathbb{Z}∕4\mathbb{Z})}^{\times }=\{1,-1\},$$

  since the only non trivial homomorphism $\mathbb{Z}∕2\mathbb{Z}\to \{1,-1\}$ maps $1$ on $-1$. The corresponding homomorphism $\phi$ is

  $$\phi \{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}(\mathbb{Z}∕4\mathbb{Z})\hfill \\ \hfill k\hfill & \hfill \mapsto \hfill & {\phi }_k\{\begin{array}{ccc}\hfill \mathbb{Z}∕4\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕4\mathbb{Z}\hfill \\ \hfill h\hfill & \hfill \mapsto \hfill & k\cdot h={(-1)}^{k}h=\{\begin{array}{cc}h\hfill & \text{ if }k=0,\hfill \\ -h\hfill & \text{ if }k=1.\hfill \end{array}\hfill \end{array}\hfill \end{array}$$

  so that ${\phi }_0=\mathrm{id}$ and ${\phi }_1=\mathrm{neg}:h\mapsto -h$.

  Then the law on $\mathbb{Z}∕4\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$ is given by

  $$(h,k)(h^{\prime },k^{\prime })=(h+{(-1)}^k{h}^{\prime },k+k^{\prime })(h,h^{\prime }\in \mathbb{Z}∕4\mathbb{Z},k,k^{\prime }\in \mathbb{Z}∕2\mathbb{Z}).$$

  Consider now the map

  $$\chi \{\begin{array}{ccc}\hfill \mathbb{Z}∕4\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & D_8\hfill \\ \hfill (h,k)\hfill & \hfill \mapsto \hfill & r^h{s}^k\hfill \end{array}$$
  • Since $|r|=4$ and $|s|=2$, $\chi$ is well defined.
  • Every element $x$ of $D_8$ is of the form $r^h{s}^k$, where $0\le h<4,0\le k<2$, so $\chi$ is surjective.
  • Since $|\mathbb{Z}∕4\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}|=8=|D_8|$, $\chi$ is a bijection.

  • We prove that $\chi$ is a homomorphism. Since $\mathit{𝑠𝑟}{s}^{-1}=r^{-1}$, then $s^{m}r{s}^{-k}=r^{{(-1)}^m}$ for all integers $m$, thus $s^m{r}^n{s}^{-m}={(s^{m}r{s}^{-m})}^n=r^{{(-1)}^{m}n}$ for all integers $n,m$, and so

    $$s^m{r}^n=r^{{(-1)}^{m}n}{s}^m(n,m\in Z).$$

    Therefore, if $(h,k),(h^{\prime },k^{\prime })\in \mathbb{Z}∕4\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$, then

    $$\begin{array}{llll}\hfill \chi (h,k)\chi (h^{\prime },k^{\prime })& =r^h{s}^k{r}^{h^{\prime }}{s}^{k^{\prime }}& \hfill & \\ \hfill & =r^h{r}^{{(-1)}^k{h}^{\prime }}{s}^k{s}^{k^{\prime }}& \hfill & \\ \hfill & =r^{h+{(-1)}^k{h}^{\prime }}{s}^{k+k^{\prime }}& \hfill & \\ \hfill & =\chi (h+{(-1)}^k{h}^{\prime },k+k^{\prime })& \hfill & \\ \hfill & =\chi ((h,k)\cdot (h^{\prime },k^{\prime })),& \hfill & \end{array}$$

    so $\chi$ is a homomorphism.

  This shows that $\chi$ is an isomorphism, so

  $$\mathbb{Z}∕4\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}\simeq D_8.$$

• Construction of $(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}){⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$.

  We know that

  $$\mathrm{Aut}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\simeq {\mathrm{GL}}_2({𝔽}_2),$$

  where ${\mathrm{GL}}_2({𝔽}_2)$ is a non abelian group of order $6$, so ${\mathrm{GL}}_2({𝔽}_2)\simeq S_3$.

  $${\mathrm{GL}}_2({𝔽}_2)=\left\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)\right\},$$

  where the first three matrices form a subgroup of order $3$, and the last three have order $2$. Put $H=\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ and $K=\mathbb{Z}∕2\mathbb{Z}$, and let $\phi :K\to \mathrm{Aut}(H)$ a non trivial homomorphism. Then $\phi$ maps $k=1$ of order $2$ on one of the three matrices $M$ of order $2$ in ${\mathrm{GL}}_2({𝔽}_2)$, thus

  $${\phi }_l\left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right)=M^l\left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right),M\in \left\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)\right\},$$

  Therefore there are exactly $3$ non trivial homomorphism $\phi ,\lambda ,\xi$, where

  $$\phi \{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\hfill \\ \hfill l\hfill & \hfill \mapsto \hfill & {\phi }_l\{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\hfill \\ \hfill \left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right)\hfill & \hfill \mapsto \hfill & {\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)}^l\left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right)\hfill \end{array}\hfill \end{array}$$

  and similar description for $\lambda ,\xi$, with $\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$ or $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)$ instead of $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$.

  The corresponding law for $\phi$ is given by

  $$\left(\left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right),l\right)\left(\left(\begin{array}{c}\hfill h^{\prime }\hfill \\ \hfill k^{\prime }\hfill \end{array}\right),l^{\prime }\right)=\left(\left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right)+{\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)}^l\left(\begin{array}{c}\hfill h^{\prime }\hfill \\ \hfill k^{\prime }\hfill \end{array}\right),l+l^{\prime }\right),$$

  or, in a more compact way,

  $$\begin{array}{lll}\hfill (h,k,l)(h^{\prime },k^{\prime },l^{\prime })=(h+h^{\prime }+l{k}^{\prime },k+k^{\prime },l+l^{\prime })(h,k,l,h^{\prime },k^{\prime },l^{\prime }\in \mathbb{Z}∕2\mathbb{Z}).& & \hfill \text{(1)}\end{array}$$

  The identity of $G$ is $1=(0,0,0)$.

  We check that this group $G=(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}){⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$ is isomorphic to $D_8$.

  Put $\rho =(1,1,1),\sigma =(0,0,1)$ in $G$. Then, using (1), ${\sigma }^2=(0,0,0)=1$ and

  $$\begin{array}{llll}\hfill {\rho }^2& =(1,1,1)(1,1,1)=(1,0,0)\ne 1,& \hfill & \\ \hfill {\rho }^3& =(1,0,0,)(1,1,1)=(0,1,1,)& \hfill & \\ \hfill {\rho }^4& =(1,0,0)(1,0,0)=(0,0,0)=1.& \hfill & \end{array}$$

Moreover,

so

Since $G=⟨(1,0,0),(0,1,0),(0,0,1)⟩$ and $(1,0,0)={\rho }^2,(0,1,0)=\mathit{𝜎𝜌},(0,0,1)=\sigma$, we obtain

Therefore there is a surjective homomorphism $\zeta :D_8=⟨r,s\mid r^4=s^2=1,\mathit{𝑠𝑟}{s}^{-1}=r^{-1}⟩\to G$. Since $|G|=|D_8|=8$, $\zeta$ is an isomorphism, and so

By the text p. 184 and Exercise 6, the groups using the homomorphisms $\psi$ and $\chi$ are isomorphic to $G$. □