Problem 5.5.16 (Semidirect products $Z_2 \rtimes_\varphi Z_8$)

Proof. We use additive notations: we count the homomorphisms from $\mathbb{Z}∕2\mathbb{Z}$ into $\mathrm{Aut}(\mathbb{Z}∕8\mathbb{Z})$.

We know that $\mathrm{Aut}(\mathbb{Z}∕8\mathbb{Z})\simeq {(\mathbb{Z}∕8\mathbb{Z})}^{\times }=\{1,3,5,7\}.$ Since every element $\alpha$ of ${(\mathbb{Z}∕8\mathbb{Z})}^{\times }$ has order $1$ or $2$, there are exactly $4$ homomorphisms from $Z_2$ into $\mathrm{Aut}(Z_8)$, given by

Put $G_{\alpha }=\mathbb{Z}∕8\mathbb{Z}{⋊}_{{\phi }_{\alpha }}\mathbb{Z}∕2\mathbb{Z}$.

• If $\alpha =1$, then ${\phi }_1$ is the trivial homomorphism, and

  $$G_1=\mathbb{Z}∕8\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\simeq Z_8\times Z_2.$$

  is abelian.

• if $\alpha =-1$, then

  $$G_{-1}=\mathbb{Z}∕8\mathbb{Z}{\times }_{{\phi }_{-1}}\mathbb{Z}∕2\mathbb{Z},$$

  where

  $${\phi }_{-1}\{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}(\mathbb{Z}∕8\mathbb{Z})\hfill \\ \hfill \bar{x}={[x]}_2\hfill & \hfill \mapsto \hfill & {\phi }_{-1}(\bar{x})\{\begin{array}{ccc}\hfill \mathbb{Z}∕8\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕8\mathbb{Z}\hfill \\ \hfill z\hfill & \hfill \mapsto \hfill & {(-1)}^{x}z\hfill \end{array}\hfill \end{array}$$

  The law on $G_{-1}$ is given by

  $$(h,k)(h^{\prime },k^{\prime })=(h+{(-1)}^k{h}^{\prime },k+k^{\prime }).$$

  Note that $r=(1,0),s=(0,1)$ satisfy $r^8=s^2=1,\mathit{𝑠𝑟}{s}^{-1}=r^{-1}$

  As in Exercise 9, we obtain the presentation of $G_{-1}$:

  $$G_{-1}\simeq ⟨r,s\mid r^8=s^2=1,\mathit{𝑠𝑟}{s}^{-1}=r^{-1}⟩,$$

  so

  $$G_{-1}\simeq D_{16}.$$

• If $\alpha =3$, then

  $$G_{-1}=\mathbb{Z}∕8\mathbb{Z}{\times }_{{\phi }_3}\mathbb{Z}∕2\mathbb{Z},$$

  where

  $${\phi }_3\{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}(\mathbb{Z}∕8\mathbb{Z})\hfill \\ \hfill \bar{x}={[x]}_2\hfill & \hfill \mapsto \hfill & {\phi }_3(\bar{x})\{\begin{array}{ccc}\hfill \mathbb{Z}∕8\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕8\mathbb{Z}\hfill \\ \hfill z\hfill & \hfill \mapsto \hfill & 3^{x}z\hfill \end{array}\hfill \end{array}$$

  The law on $G_{-1}$ is given by

  $$(h,k)(h^{\prime },k^{\prime })=(h+3^k{h}^{\prime },k+k^{\prime }).$$

  Note that $\sigma =(1,0),\tau =(0,1)$ satisfy ${\sigma }^8={\tau }^2=1$, and

  $$\mathit{𝜏𝜎}{\tau }^{-1}=(0,1)(1,0){(0,1)}^{-1}=({\phi }_3(1)(1),0)=(3,0)={\sigma }^3.$$

  As in Exercise 9, we obtain the presentation of $G_3$:

  $$G_3\simeq ⟨\sigma ,\tau \mid {\sigma }^8={\tau }^2=1,\mathit{𝜎𝜏}=\tau {\sigma }^3⟩,$$

  so by the definition of $Q{D}_{16}$ in Exercise 2.5.11,

  $$G_3\simeq Q{D}_{16}.$$

• If $\alpha =-3$, then

  $$G_{-1}=\mathbb{Z}∕8\mathbb{Z}{\times }_{{\phi }_{-3}}\mathbb{Z}∕2\mathbb{Z},$$

  where

  $${\phi }_{-3}\{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}(\mathbb{Z}∕8\mathbb{Z})\hfill \\ \hfill \bar{x}={[x]}_2\hfill & \hfill \mapsto \hfill & {\phi }_{-3}(\bar{x})\{\begin{array}{ccc}\hfill \mathbb{Z}∕8\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕8\mathbb{Z}\hfill \\ \hfill z\hfill & \hfill \mapsto \hfill & {(-3)}^{x}z\hfill \end{array}\hfill \end{array}$$

  The law on $G_{-3}$ is given by

  $$(h,k)(h^{\prime },k^{\prime })=(h+{(-3)}^k{h}^{\prime },k+k^{\prime }).$$

  Note that $u=(1,0),v=(0,1)$ satisfy $u^8=v^2=1$, and

  $$\mathit{𝑢𝑣}{u}^{-1}=(0,1)(1,0){(0,1)}^{-1}=({\phi }_{-3}(1)(1),0)=(-3,0)=u^{-3}=u^5.$$

  As in Exercise 9, we obtain the presentation of $G_{-3}$:

  $$G_{-3}\simeq ⟨u,v\mid u^8=v^2=1,\mathit{𝑣𝑢}=u{v}^5⟩,$$

  so by the definition of $M$ in Exercise 2.5.11,

  $$G_{-3}\simeq M,$$

  where $M$ is the modular group of order $16$.