Problem 5.5.7 (Groups of order 56)

Proof. Let $G$ be a group of order $56=2^3\cdot 7$.

(a)

If $G$ is abelian, by Theorem 5 of Chapter 5, there are three isomorphism types of abelian groups of order $2^3\cdot 7$: $$G\simeq Z_{2^3}\times Z_7\text{or}G\simeq Z_{2^2}\times Z_2\times Z_7\text{or}\text{or}G\simeq Z_2\times Z_2\times Z_2\times Z_7$$

(b)

By Sylow’s Theorem, if $n_p$ denotes the number of Sylow $p$-subgroups of $G$, then $$\begin{array}{lll}\hfill n_2\mid 7,n_2\equiv 1(\mathit{𝑚𝑜𝑑}2),& & \hfill \\ \hfill n_7\mid 8,n_7\equiv 1(\mathit{𝑚𝑜𝑑}7),& & \hfill \end{array}$$

so

$$n_2\in \{1,7\},n_7\in \{1,8\}.$$

Assume $n_7=8$. Since the $7$-Sylow subgroups have trivial intersection, these subgroups contain one element of order 1 and $8\cdot 6=48$ element of order $7$. It remains only $7$ elements, so there is at most one Sylow $2$-subgroup of order $8=1+7$, so $n_2=1$. Therefore

$$n_2=1\text{or}{n}_7=1.$$

Every group of order $56$ has either a normal Sylow $2$-subgroup or a normal Sylow $7$-subgroup.

(c)

Suppose here that the non-abelian group $G$ has a normal Sylow $7$-subgroup $H$ ( $n_7=1$), and let $S$ be a Sylow $2$-subgroup of $G$. The order of an element $x\in H\cap S$ divides $7$ and $8$, thus $x=1$, so $H\cap S=\{1\}$. Since $H⊴G$, $\mathit{𝐻𝑆}$ is a subgroup of $G$, and $|\mathit{𝐻𝑆}|=|H||S|∕|H\cap S|=56$, thus $G=\mathit{𝐻𝑆}$. Since $$G=\mathit{𝐻𝑆},H⊴G,H\cap S=\{1\},$$

Theorem 12 shows that

$$G\simeq H{⋊}_{\phi }S,$$

where the homomorphism $\phi :S\to \mathrm{Aut}(H)$ is associated to the action of $S$ on $H$ by conjugation. There are $5$ types of groups of order $8$, so we distinguish five cases.

• If $S\simeq Z_2\times Z_2\times Z_2$, then (in additive notations)

  $$G\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{\psi }(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}).$$

  where $\psi :\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})$ is a homomorphism.

  Since $\mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})\simeq {(\mathbb{Z}∕7\mathbb{Z})}^{\times }\simeq Z_6$, there is only one element of order $2$ in $\mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})$, given by

  $$\mathrm{neg}\{\begin{array}{ccc}\hfill \mathbb{Z}∕7\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕7\mathbb{Z}\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & -x\hfill \end{array}.$$

  The homomorphism $\psi$ maps the elements $(1,0,0),(0,1,0),(0,0,1)$ of order $2$ on an automorphism of order $1$ or $2$, so $\psi (1,0,0)\in \{\mathrm{id},\mathrm{neg}\}$, and so on: there exist ${\eta }_1,{\eta }_2,{\eta }_3\in \mathbb{Z}∕2\mathbb{Z}$ such that

  $$\psi (1,0,0)={\mathrm{neg}}^{{\eta }_1},\psi (0,1,0)={\mathrm{neg}}^{{\eta }_2},\psi (0,0,1)={\mathrm{neg}}^{{\eta }_3}.$$

  Then $\psi (a,b,c)=\psi {(1,0,0)}^a\psi {(0,1,0)}^b\psi {(0,0,1)}^c$, where $a,b,c\in \mathbb{Z}∕2\mathbb{Z}$, thus $\psi (a,b,c)={\mathrm{neg}}^{{\eta }_{1}a+{\eta }_{2}b+{\eta }_{3}c}$, and so

  $$\begin{array}{lll}\hfill \psi (a,b,c)(x)={(-1)}^{{\eta }_{1}a+{\eta }_{2}b+{\eta }_{3}c}x(x\in \mathbb{Z}∕7\mathbb{Z}).& & \hfill \text{(1)}\end{array}$$

  Conversely, if $\psi$ is defined by (1), then for all $x\in \mathbb{Z}∕7\mathbb{Z}$,

  $$\begin{array}{llll}\hfill [\psi (a,b,c)\circ \psi (a^{\prime },b{,}^{\prime },c^{\prime })](x)& ={(-1)}^{{\eta }_{1}a+{\eta }_{2}b+{\eta }_{3}c}{(-1)}^{{\eta }_1{a}^{\prime }+{\eta }_2{b}^{\prime }+{\eta }_3{c}^{\prime }}x& \hfill & \\ \hfill & ={(-1)}^{{\eta }_1(a+a^{\prime })+{\eta }_2(b+b^{\prime })+{\eta }_3(c+c^{\prime })}x,& \hfill & \end{array}$$

  so $\psi (a,b,c)\circ \psi (a^{\prime },b{,}^{\prime },c^{\prime })=\psi (a+a^{\prime },b+b^{\prime }n,c+c^{\prime })=\psi ((a,b,c)+(a^{\prime },b^{\prime },c^{\prime }))$ and $\psi$ is an homomorphism. This shows that there are $8$ homomorphisms defined by (1).

  If $({\eta }_1,{\eta }_2,{\eta }_3)=(0,0,0)$ then $\psi$ is the trivial homomorphism, and $G\simeq H\times S$ is abelian. There are $7$ non trivial homomorphisms, for $({\eta }_1,{\eta }_2,{\eta }_3)\ne (0,0,0)$.

  Consider two such homomorphisms $\psi ,{\psi }^{\prime }$ defined by

  $$\begin{array}{lll}\hfill \psi (a,b,c)(x)& ={(-1)}^{{\eta }_{1}a+{\eta }_{2}b+{\eta }_{3}c}x& \hfill \text{(2)}\\ \hfill {\psi }^{\prime }(a,b,c)(x)& ={(-1)}^{{\eta }_1^{\prime }a+{\eta }_2^{\prime }b+{\eta }_3^{\prime }c}x,& \hfill \text{(3)}\end{array}$$

  where $({\eta }_1,{\eta }_2,{\eta }_3)\ne (0,0,0),({\eta }_1^{\prime },{\eta }_2^{\prime },{\eta }_3^{\prime })\ne (0,0,0)$.

  We show $\mathbb{Z}∕7\mathbb{Z}{⋊}_{\psi }(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }^{\prime }}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$, by proving ${\psi }^{\prime }=\psi \circ f$, where $f\in \mathrm{Aut}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\simeq {\mathrm{GL}}_3({𝔽}_2)$ (see “Missing Theorem: precomposition by an automorphism” in the Appendix to Exercise 6).

  We can rewrite (2) and (3) in the form

  $$\psi (u)(x)={(-1)}^{\eta (u)}x,{\psi }^{\prime }(u)(x)={(-1)}^{{\eta }^{\prime }(u)}x,$$

  where $u=\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right)\in {𝔽}_2^3$, and $\eta ={𝔽}_2^3\to {𝔽}_2$ is a linear form of matrix $\left(\begin{array}{ccc}\hfill {\eta }_1\hfill & \hfill {\eta }_2\hfill & \hfill {\eta }_3\hfill \end{array}\right)$. Since $\eta \ne 0$ and ${\eta }^{\prime }\ne 0$, there is some $M\in {\mathrm{GL}}_3({𝔽}_2)$ such that

  $$\left(\begin{array}{ccc}\hfill {\eta }_1^{\prime }\hfill & \hfill {\eta }_2^{\prime }\hfill & \hfill {\eta }_3^{\prime }\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill {\eta }_1\hfill & \hfill {\eta }_2\hfill & \hfill {\eta }_3\hfill \end{array}\right)M,$$

  so that ${\eta }^{\prime }=\eta \circ f$, where $f\in \mathrm{Aut}({𝔽}_2^3)$ has matrix $M$. Then ${\psi }^{\prime }(u)(x)={(-1)}^{{\eta }^{\prime }(u)}x={(-1)}^{\eta (f(u))}x$ for all $x\in \mathbb{Z}∕7\mathbb{Z}$, so that ${\psi }^{\prime }=\psi \circ f$.

  This shows $\mathbb{Z}∕7\mathbb{Z}{⋊}_{\psi }(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }^{\prime }}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$.

  In conclusion, $G\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{\psi }(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$, where $\psi$ is any of the $7$ previous non trivial homomorphisms, for instance

  $$\psi (a,b,c)(x)={(-1)}^{a}x(x\in \mathbb{Z}∕7\mathbb{Z}).$$

  Then the law on $\mathbb{Z}∕7\mathbb{Z}{⋊}_{\psi }(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$ is given by

  $$(x,(a,b,c))(x^{\prime },(a^{\prime },b^{\prime },c^{\prime }))=(x+{(-1)}^a{x}^{\prime },(a+a^{\prime },b+b^{\prime },c+c^{\prime })),$$

  where $x,x^{\prime }\in \mathbb{Z}∕7\mathbb{Z},(a,b,c),(a^{\prime },b^{\prime },c^{\prime })\in {(\mathbb{Z}∕2\mathbb{Z})}^3$.

  There is only one type of non abelian groups of order $56$ which have a normal Sylow $7$-subgroup and whose Sylow $2$-subgroup is isomorphic to $Z_2\times Z_2\times Z_2$.

• If $S\simeq Z_4\times Z_2$, then

  $$G\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{\psi }(\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}).$$

  where $\psi :\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})$ is a homomorphism. Since $\psi {(1,0)}^4=1$, $|\psi (1,0)|$ divides $4$, and so $\psi (1,0)\in \{\mathrm{id},\mathrm{neg}\}$, and similarly since $\psi {(0,1)}^2=\mathrm{id}$, $\psi (0,1)\in \{\mathrm{id},\mathrm{neg}\}$:

  $$\psi (1,0)={\mathrm{neg}}^{{\eta }_1},\psi (0,1)={\mathrm{neg}}^{{\eta }_2},({\eta }_1,{\eta }_2\in \{0,1\}).$$

  If $\bar{a}={[a]}_4\in \mathbb{Z}∕4\mathbb{Z}$ and $\bar{b}={[b]}_2\in \mathbb{Z}∕2\mathbb{Z}$, where $a,b\in \mathbb{Z}$, then

  $$\psi (\bar{a},\bar{b})=\psi {(1,0)}^a\psi {(0,1)}^b={\mathrm{neg}}^{{\eta }_{1}a+{\eta }_{2}b},$$

  so, for all $x\in \mathbb{Z}∕7\mathbb{Z}$,

  $$\begin{array}{lll}\hfill \psi (\bar{a},\bar{b})(x)={(-1)}^{{\eta }_{1}a+{\eta }_{2}b}x.& & \hfill \text{(4)}\end{array}$$

  As in the previous item, every map $\psi :Z_4\times Z_2\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})$ defined by (4) is a homomorphism.

  Therefore there are exactly $3$ non trivial homomorphisms $\psi :\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})$, defined by

  $$\begin{array}{llll}\hfill {\psi }_1(\bar{a},\bar{b})(x)& ={(-1)}^{a}x,& \hfill & \\ \hfill {\psi }_2(\bar{a},\bar{b})(x)& ={(-1)}^{b}x,& \hfill & \\ \hfill {\psi }_3(\bar{a},\bar{b})(x)& ={(-1)}^{a+b}x.& \hfill & \end{array}$$

  Consider the automorphism $r\in \mathrm{Aut}(\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$ (see the appendix 1) characterized by

  $$r\{\begin{array}{ccc}\hfill \alpha \hfill & \hfill \mapsto \hfill & \hfill \beta \hfill \\ \hfill \beta \hfill & \hfill \mapsto \hfill & \hfill -\alpha \hfill \end{array}$$

  where $\alpha =(1,0)$ and $\beta =(1,1)$ are generators of $\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$.

  Since $(\bar{a},\bar{b})=(a-b)(1,0)+b(1,1)=(a-b)\alpha +\mathit{𝑏𝛽}$,

  $$\begin{array}{llll}\hfill r(\bar{a},\bar{b})& =(a-b)r(\alpha )+br(\beta )& \hfill & \\ \hfill & =-(a-b)\beta -\mathit{𝑏𝛼}& \hfill & \\ \hfill & =(a-b)(1,1)-b(1,0)& \hfill & \\ \hfill & =(-\bar{a}+2\bar{b},-\bar{a}+\bar{b})& \hfill & \\ \hfill & =(\bar{a}-2\bar{b},\bar{a}-\bar{b})& \hfill & \\ \hfill & =(\bar{a}+2\bar{b},\bar{a}+\bar{b}).& \hfill & \end{array}$$

  Then, for all $(\bar{a},\bar{b})\in \mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$, and for all $x\in \mathbb{Z}∕7\mathbb{Z}$,

  $$\begin{array}{llll}\hfill {\psi }_2(r(\bar{a},\bar{b}))(x)& ={\psi }_2(\overline{a+2b},\overline{a+b})(x)& \hfill & \\ \hfill & ={(-1)}^{a+b}x& \hfill & \\ \hfill & ={\psi }_3(\bar{a},\bar{b})(x).& \hfill & \end{array}$$

  This gives

  $${\psi }_2={\psi }_3\circ r,\text{where }r\in \mathrm{Aut}(\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}).$$

  By the missing Theorem “precomposition by an automorphism” in the Appendix to Exercise 6,

  $$\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_2}(\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_3}(\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}).$$

  Therefore

  $$G\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_1}(\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\text{ or }G\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_2}(\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}),$$

  where

  $$\begin{array}{llll}\hfill {\psi }_1(\bar{a},\bar{b})(x)& ={(-1)}^{a}x,& \hfill & \\ \hfill {\psi }_2(\bar{a},\bar{b})(x)& ={(-1)}^{b}x,& \hfill & \\ \hfill & & \hfill \end{array}$$

  The laws of these two groups are defined by

  $$\begin{array}{lll}\hfill (x,(a,b))(x^{\prime },(a^{\prime },b^{\prime }))=(x+{(-1)}^a{x}^{\prime },(a+a^{\prime },b+b^{\prime })),& & \hfill \\ \hfill (x,(a,b))(x^{\prime },(a^{\prime },b^{\prime }))=(x+{(-1)}^b{x}^{\prime },(a+a^{\prime },b+b^{\prime }))& & \hfill \end{array}$$

  $(x,x^{\prime }\in \mathbb{Z}∕7\mathbb{Z},a,a^{\prime }\in \mathbb{Z}∕4\mathbb{Z},b,b^{\prime }\in \mathbb{Z}∕2\mathbb{Z}).$

  It remains to verify that these two groups are not isomorphic, by computing the kernels of ${\psi }_1$ and ${\psi }_2$. If $(\bar{a},\bar{b})\in \mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$,

  $$\begin{array}{llll}\hfill {\psi }_1(\bar{a},\bar{b})=\mathrm{id}& ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z}∕7\mathbb{Z},{(-1)}^{a}x=x& \hfill & \\ \hfill & ⟺{(-1)}^a=1& \hfill & \\ \hfill & ⟺\bar{a}\in \{0,2\}& \hfill & \\ \hfill & ⟺(\bar{a},\bar{b})\in \{(0,0),(0,1),(2,0),(2,2)\}.& \hfill & \end{array}$$

  So

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->({\psi }_1)=\{(0,0),(0,1),(2,0),(2,2)\},$$

  Since all elements have order $2$,

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->({\psi }_1)\simeq Z_2\times Z_2.$$

  Similarly,

  $$\begin{array}{llll}\hfill {\psi }_2(\bar{a},\bar{b})=\mathrm{id}& ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z}∕7\mathbb{Z},{(-1)}^{b}x=x& \hfill & \\ \hfill & ⟺{(-1)}^b=1& \hfill & \\ \hfill & ⟺\bar{b}=0& \hfill & \\ \hfill & ⟺(\bar{a},\bar{b})\in \{(0,0),(1,0),(2,0),(3,0)\}=⟨(1,0)⟩.& \hfill & \end{array}$$

  Therefore

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->({\psi }_2)\simeq {\mathbb{Z}}_4.$$

  Since $\ker <!--\; FUNCTION\; APPLICATION\; -->({\psi }_1)$ and $\ker <!--\; FUNCTION\; APPLICATION\; -->({\psi }_2)$ are not isomorphic, $\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_1}(\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$ and $\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_2}(\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$ are not isomorphic by Theorem A given in the Appendix 2.

  There are two types of non abelian groups of order $56$ which have a normal Sylow $7$-subgroup and whose Sylow $2$-subgroup is isomorphic to $Z_4\times Z_2$.

• If $S\simeq Z_8$, then

  $$G\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{\psi }\mathbb{Z}∕8\mathbb{Z}.$$

  where $\psi :\mathbb{Z}∕8\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})\simeq Z_6$ is a homomorphism.

  Since $1$ has order $8$ in $\mathbb{Z}∕8\mathbb{Z}$, then $\psi {(1)}^8=\mathrm{id}$, thus $\psi {(1)}^2=\mathrm{id}$ and $\psi (1)\in \{\mathrm{id},\mathrm{neg}\}$. Therefore $\psi (1)={\mathrm{neg}}^{\eta }$ for some $\eta \in \{0,1\}$, and for all $x\in \mathbb{Z}∕7\mathbb{Z}$,

  $$\psi (\bar{a})(x)={(-1)}^{\mathit{𝜂𝑎}}x,(a\in \mathbb{Z},\eta \in \{0,1\}).$$

  There is only one non trivial homomorphism ${\psi }_0$, given by

  $${\psi }_0(\bar{a})(x)={(-1)}^{a}x.$$

  Therefore

  $$G\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_0}\mathbb{Z}∕8\mathbb{Z},$$

  where ${\psi }_0$ is given by

  $${\psi }_0(\bar{a})(x)={(-1)}^{a}x.$$

  The law on $\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_0}\mathbb{Z}∕8\mathbb{Z}$ is defined by

  $$(x,y)(x^{\prime },y^{\prime })=(x+{(-1)}^y{x}^{\prime },y+y^{\prime }).$$

  There is only one type of non abelian groups of order $56$ which have a normal Sylow $7$-subgroup and whose Sylow $2$-subgroup is isomorphic to $Z_8$.

• If $S\simeq Q_8$, then

  $$G\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{\psi }{Q}_8.$$

  where $\psi :Q_8\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})\simeq Z_6$ is a homomorphism.

  By Exercise 6.3.7,

  $$Q_8=⟨i,j\mid i^2=j^2,i^{-1}\mathit{𝑗𝑖}=j^{-1}⟩.$$

  Since $i^4=j^4=1$, then $\psi {(i)}^4=\psi {(j)}^4=1$, thus $\psi (i)\in \{\mathrm{id},\mathrm{neg}\}$ and $\psi (j)\in \{\mathrm{id},\mathrm{neg}\}$, so $\psi (i)={\mathrm{neg}}^{{\eta }_1}$ and $\psi (j)={\mathrm{neg}}^{{\eta }_2}$, for some ${\eta }_1,{\eta }_2\in \{0,1\}$.

  Conversely, if ${\eta }_1,{\eta }_2\in \{0,1\}$, then $I={\mathrm{neg}}^{{\eta }_1}$ and $J={\mathrm{neg}}^{{\eta }_2}$ satisfy $I^2=J^2(=\mathrm{id})$ and $I^{-1}\mathit{𝐽𝐼}={\mathrm{neg}}^{-{\eta }_1}{\mathrm{neg}}^{{\eta }_2}{\mathrm{neg}}^{{\eta }_1}={\mathrm{neg}}^{{\eta }_2}=J=J^{-1}$. Therefore $I$ and $J$ satisfy the relations of the presentation of $Q_8$, so there is a homomorphism $\psi :Q_8\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})$ such that $\psi (i)=I={\mathrm{neg}}^{{\eta }_1}$ and $\psi (j)=J={\mathrm{neg}}^{{\eta }_2}$.

  In conclusion, there are four homomorphisms $\psi :Q_8\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})$ characterized by

  $$\begin{array}{lll}\hfill \{\begin{array}{c}{\psi }_0(i)(x)=x\hfill \\ {\psi }_0(j)(x)=x\hfill \end{array}\{\begin{array}{c}{\psi }_1(i)(x)=-x\hfill \\ {\psi }_1(j)(x)=x\hfill \end{array}\{\begin{array}{c}{\psi }_2(i)(x)=x\hfill \\ {\psi }_2(j)(x)=-x\hfill \end{array}\{\begin{array}{c}{\psi }_3(i)(x)=-x\hfill \\ {\psi }_3(j)(x)=-x\hfill \end{array}& & \hfill \end{array}$$

  for all $x\in \mathbb{Z}∕7\mathbb{Z}$ .

  Then ${\psi }_0$ is the trivial homomorphism, and the corresponding group $G_0$ is the direct product $G_0\simeq \mathbb{Z}∕7\mathbb{Z}\times Q_8$, which is not abelian.

  Explicitly, ${\psi }_1,{\psi }_2,{\psi }_3$ are given by

  $y$	$1$	$-1$	$i$	$-i$	$j$	$-j$	$k$	$-k$
  ${\psi }_1(y)$	$\mathrm{id}$	$\mathrm{id}$	$\mathrm{neg}$	$\mathrm{neg}$	$\mathrm{id}$	$\mathrm{id}$	$\mathrm{neg}$	$\mathrm{neg}$

  
  

  $x$	$1$	$-1$	$i$	$-i$	$j$	$-j$	$k$	$-k$
  ${\psi }_2(x)$	$\mathrm{id}$	$\mathrm{id}$	$\mathrm{id}$	$\mathrm{id}$	$\mathrm{neg}$	$\mathrm{neg}$	$\mathrm{neg}$	$\mathrm{neg}$

  
  

  $x$	$1$	$-1$	$i$	$-i$	$j$	$-j$	$k$	$-k$
  ${\psi }_3(x)$	$\mathrm{id}$	$\mathrm{id}$	$\mathrm{neg}$	$\mathrm{neg}$	$\mathrm{neg}$	$\mathrm{neg}$	$\mathrm{id}$	$\mathrm{id}$

  By the presentation of $Q_8$, since $j$ and $i$ in place of $i$ and $j$ satisfy the relations of the presentation of $Q_8$, there is a unique automorphism $f\in \mathrm{Aut}(Q_8)$ such that $f(i)=j$ and $f(j)=i$. Then

  $$({\psi }_1\circ f)(i)={\psi }_1(j)=\mathrm{id}={\psi }_2(i),({\psi }_1\circ f)(j)={\psi }_1(i)=\mathrm{neg}={\psi }_2(j).$$

  Therefore ${\psi }_2={\psi }_1\circ f$, and so, by the missing Theorem “precomposition by an automorphism” in the Appendix to Exercise 6,

  $$\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_1}{Q}_8\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_2}{Q}_8.$$

  Similarly, there exists a unique $g\in \mathrm{Aut}(Q_8)$ such that $g(i)=k$ and $g(j)=i$. Then

  $$({\psi }_1\circ g)(i)={\psi }_1(k)=\mathrm{neg}={\psi }_3(i),({\psi }_1\circ g)(j)={\psi }_1(i)=\mathrm{neg}={\psi }_3(j).$$

  Therefore ${\psi }_3={\psi }_1\circ g$, and so

  $$\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_1}{Q}_8\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_3}{Q}_8.$$

  Moreover, $G_0\not\simeq G_1$ since the kernel of the trivial homomorphism ${\psi }_0$ is $Q_8$, and $\ker <!--\; FUNCTION\; APPLICATION\; -->({\psi }_1)=\{1,-1,j,-j\}\not\simeq Q_8$.

  The laws are defined by

  $$\begin{array}{llll}\hfill G_0:& (a,x)(a^{\prime },x^{\prime })=(a+a^{\prime },x{x}^{\prime }),& \hfill & \\ \hfill G_1:& (a,x)(a^{\prime },x^{\prime })=(a+𝜀a^{\prime },x{x}^{\prime }),𝜀=-1\text{ if }x\in \{\pm i,\pm k\},𝜀=1\text{ otherwise}.& \hfill & \end{array}$$

  There are two types of non abelian groups of order $56$ which have a normal Sylow $7$-subgroup and whose Sylow $2$-subgroup is isomorphic to $Q_8$.

• If $S\simeq D_8$, then

  $$G\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{\psi }{D}_8.$$

  where $\psi :D_8\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})\simeq Z_6$ is a homomorphism.

  By definition, $D_8=⟨r,s⟩$, where $r^4=s^2=1$ and $\mathit{𝑠𝑟}{s}^{-1}=r^{-1}$, and every element $x\in D_8$ is of the form $x=r^k{s}^l$ for some $k\in \{0,1,2,3\}$ and some $l\in \{0,1\}$.

  Since $\psi$ is a homomorphism, $\psi {(r)}^4=\mathrm{id}$ and $\psi {(r)}^2=\mathrm{id}$, thus $\psi (r)\in \{\mathrm{id},\mathrm{neg}\}$ and $\psi (r)\in \{\mathrm{id},\mathrm{neg}\}$, so $\psi (r)={\mathrm{neg}}^{{\eta }_1}$ and $\psi (s)={\mathrm{neg}}^{{\eta }_2}$, for some ${\eta }_1,{\eta }_2\in \{0,1\}$. For all $x\in \mathbb{Z}∕7\mathbb{Z}$,

  $$\psi (r)(x)={(-1)}^{{\eta }_1}x,\psi (s)(x)={(-1)}^{{\eta }_2}x,$$

  therefore, for all $x\in \mathbb{Z}∕7\mathbb{Z}$, and for $0\le k<4,0\le l<2$,

  $$\begin{array}{lll}\hfill \psi (r^k{s}^l)(x)={(-1)}^{{\eta }_{1}k+{\eta }_{2}l}x({\eta }_1,{\eta }_2\in \{0,1\}).& & \hfill \text{(5)}\end{array}$$

  Conversely, if $\psi$ is defined by (5) for some ${\eta }_1,{\eta }_2\in \{0,1\}$, then $\psi :\mathbb{Z}∕8\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})$ is a homomorphism: using $(s^l{r}^k)(s^{l^{\prime }}{r}^{k^{\prime }})=r^{k+{(-1)}^l{k}^{\prime }}{s}^{l+l^{\prime }}$, we obtain for $x\in \mathbb{Z}∕7\mathbb{Z}$,

  $$\begin{array}{llll}\hfill \psi (r^k{s}^l)(r^{k^{\prime }}{s}^{l^{\prime }})(x)& =\psi (r^{k+{(-1)}^l{k}^{\prime }}{s}^{l+l^{\prime }})x& \hfill & \\ \hfill & ={(-1)}^{{\eta }_1(k+{(-1)}^l{k}^{\prime })+{\eta }_2(l+l^{\prime })}x& \hfill & \\ \hfill [\psi (r^k{s}^l)\psi (r^{k^{\prime }}{s}^{l^{\prime }})](x)& ={(-1)}^{{\eta }_1(k+k^{\prime })+{\eta }_2(l+l^{\prime })}x.& \hfill & \end{array}$$

  Since ${(-1)}^{{\eta }_1{(-1)}^l{k}^{\prime }}={(-1)}^{{\eta }_1{k}^{\prime }}$ (because ${(-1)}^u={(-1)}^{-u}$ for all $u$), we obtain $\psi (r^k{s}^l)(r^{k^{\prime }}{s}^{l^{\prime }})=\psi (r^k{s}^l)\psi (r^{k^{\prime }}{s}^{l^{\prime }})$, so $\psi$ is a homomorphism.

  (Alternatively, this is a consequence of the presentation of $D_8$.)

  Therefore there are $4$ homomorphisms $\psi :\mathbb{Z}∕8\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})$, defined, for $x\in \mathbb{Z}∕7\mathbb{Z}$, by

  $$\begin{array}{llll}\hfill {\psi }_0(r^k{s}^l)(x)& =x,& \hfill & \\ \hfill {\psi }_1(r^k{s}^l)(x)& ={(-1)}^{k}x,& \hfill & \\ \hfill {\psi }_2(r^k{s}^l)(x)& ={(-1)}^{l}x,& \hfill & \\ \hfill {\psi }_3(r^k{s}^l)(x)& ={(-1)}^{k+l}x.& \hfill & \end{array}$$

  Since ${\psi }_0$ is the trivial homomorphism, the corresponding group $G_0=\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_0}{D}_8$ is the non abelian direct product $\mathbb{Z}∕7\mathbb{Z}\times D_8$.

  Since $D_8\simeq \mathbb{Z}∕4\mathbb{Z}⋊\mathbb{Z}∕2\mathbb{Z}$, we can write the laws of these tree groups $G_1,G_2,G_3$ in the form

  $$\begin{array}{llll}\hfill G_1:(x,k,l)(x^{\prime },k^{\prime },l^{\prime })& =(x+{(-1)}^k{x}^{\prime },k+{(-1)}^l{k}^{\prime },l+l^{\prime }),& \hfill & \\ \hfill G_2:(x,k,l)(x^{\prime },k^{\prime },l^{\prime })& =(x+{(-1)}^l{x}^{\prime },k+{(-1)}^l{k}^{\prime },l+l^{\prime }),& \hfill & \\ \hfill G_3:(x,k,l)(x^{\prime },k^{\prime },l^{\prime })& =(x+{(-1)}^{k+l}{x}^{\prime },k+{(-1)}^l{k}^{\prime },l+l^{\prime }),& \hfill & \\ \hfill & & \hfill \end{array}$$

  where $(x,k,l),(x^{\prime },k^{\prime },l^{\prime })\in \mathbb{Z}∕7\mathbb{Z}\times \mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$.

  We compute the kernels of ${\psi }_1,{\psi }_2,{\psi }_3$:

  $$\begin{array}{llll}\hfill r^k{s}^l\in \ker <!--\; FUNCTION\; APPLICATION\; -->{\psi }_1& ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z}∕7\mathbb{Z},{(-1)}^{k}x=x& \hfill & \\ \hfill & ⟺{(-1)}^k=1& \hfill & \\ \hfill & ⟺r^k{s}^l\in \{1,s,r^2,r^{2}s\}.& \hfill & \end{array}$$

  So

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->({\psi }_1)=\{1,s,r^2,r^{2}s\}\simeq Z_2\times Z_2.$$

  Similarly

  $$\begin{array}{llll}\hfill r^k{s}^l\in \ker <!--\; FUNCTION\; APPLICATION\; -->{\psi }_2& ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z}∕7\mathbb{Z},{(-1)}^{l}x=x& \hfill & \\ \hfill & ⟺{(-1)}^l=1& \hfill & \\ \hfill & ⟺r^k{s}^l\in \{1,r,r^2,r^3\}.& \hfill & \end{array}$$

  So

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->({\psi }_2)=⟨r⟩\simeq Z_4.$$

  Finally,

  $$\begin{array}{llll}\hfill r^k{s}^l\in \ker <!--\; FUNCTION\; APPLICATION\; -->{\psi }_3& ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z}∕7\mathbb{Z},{(-1)}^{k+l}x=x& \hfill & \\ \hfill & ⟺{(-1)}^{k+l}=1& \hfill & \\ \hfill & ⟺r^k{s}^l\in \{1,\mathit{𝑟𝑠},r^2,r^{3}s\}.& \hfill & \end{array}$$

  So

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->({\psi }_3)=\{1,\mathit{𝑟𝑠},r^2,r^{3}s\}\simeq Z_2\times Z_2.$$

  This shows that $\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_1}{D}_8$ is not isomorphic to $\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_2}{D}_8$, but we cannot conclude for ${\psi }_3$. Note that ${\psi }_1$ and ${\psi }_3$ are characterized by

  $$\begin{array}{llll}\hfill & {\psi }_1(r)(x)=-x,{\psi }_1(s)(x)=x,& \hfill & \\ \hfill & {\psi }_3(r)(x)=-x,{\psi }_3(s)(x)=-x(x\in \mathbb{Z}∕7\mathbb{Z}).& \hfill & \end{array}$$

  Since

  $$\begin{array}{llll}\hfill & r^4={(\mathit{𝑟𝑠})}^2=1,& \hfill & \\ \hfill & (\mathit{𝑟𝑠})r{(\mathit{𝑟𝑠})}^{-1}=r(\mathit{𝑠𝑟}{s}^{-1})r^{-1}=r{r}^{-1}{r}^{-1}=r^{-1},& \hfill & \end{array}$$

  the ordered pair $(r,\mathit{𝑟𝑠})$ satisfy the relation of the presentation of $D_8$:

  $$D_8=⟨r,s\mid r^4=s^2,\mathit{𝑠𝑟}{s}^{-1}=r^{-1}\}.$$

  Therefore there is a unique homomorphism $f:D_8\to D_8$ such that $f(r)=r$ and $f(s)=\mathit{𝑟𝑠}$. Since $D_8=⟨r,\mathit{𝑟𝑠}⟩$, $f:D_8\to D_8$ is surjective, where $D_8$ is finite, so $f$ is an automorphism. Moreover, for all $x\in \mathbb{Z}∕7\mathbb{Z}$,

  $$\begin{array}{llll}\hfill ({\psi }_1\circ f)(r)(x)& ={\psi }_1(r)(x)=-x,& \hfill & \\ \hfill ({\psi }_1\circ f)(s)(x)& ={\psi }_1(\mathit{𝑟𝑠})(x)={\psi }_1(r)({\psi }_1(s))(x))=-x.& \hfill & \end{array}$$

  Therefore ${\psi }_1\circ f={\psi }_3$, where $f\in \mathrm{Aut}(D_8)$, and so, by the precomposition by an automorphism in the Appendix to Exercise 6,

  $$\mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_1}{D}_8\simeq \mathbb{Z}∕7\mathbb{Z}{⋊}_{{\psi }_3}{D}_8.$$

  Moreover, since ${\psi }_0$ is the trivial homomorphism, with kernel $D_8$, $G_0\simeq \mathbb{Z}∕7\mathbb{Z}\times D_8$ is not isomorphic to $G_1$, nor to $G_2$.

  There are three types of non abelian groups of order $56$ which have a normal Sylow $7$-subgroup and whose Sylow $2$-subgroup is isomorphic to $D_8$.

(d)

Let $G$ be a group of order $56$ with a nonnormal Sylow $7$-subgroup. By part (b), the Sylow $2$-subgroup $S$ is normal in $G$ ( $n_2=1$). Since $H\cap S=\{1\}$, we obtain $$G\simeq S{⋊}_{\psi }H,H\simeq Z_7,$$

where $\psi :H\to \mathrm{Aut}(S)$ is a homomorphism. Since $|\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )|$ divides $|H|=7$, $|\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )|=1$ or $|\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )|=7$. If $|\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )|=7$, then $\psi$ is the trivial homomorphism, thus $G\simeq S\times H$, and $H⊴G$, which is in contradiction with the hypothesis. Therefore

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )=\{1\}.$$

Since $\psi$ is injective, $H\simeq \mathrm{im}(\psi )\le \mathrm{Aut}(S)$, therefore

$$7\mid |\mathrm{Aut}(S)|.$$

The group $S$ of order $8$ is isomorphic to $Z_2\times Z_2\times Z_2,Z_4\times Z_2,Z_8,D_8$ or $Q_8$. The following array gives the order of $|\mathrm{Aut}(S)|$ in each case:

$S$	$Z_2\times Z_2\times Z_2$	$Z_4\times Z_2$	$Z_8$	$D_8$	$Q_8$
$|\mathrm{Aut}(S)|$	$168=7\cdot 6\cdot 4$	$8$	$4$	$8$	$24$

Since $7$ divides $|\mathrm{Aut}(S)|$, the only possibility is

$$S\simeq Z_2\times Z_2\times Z_2.$$

(e)

By part (d) (in additive notations), $$G\simeq {(\mathbb{Z}∕2\mathbb{Z})}^3{⋊}_{\phi }\mathbb{Z}∕7\mathbb{Z},$$

where $\phi :\mathbb{Z}∕7\mathbb{Z}\to \mathrm{Aut}({(\mathbb{Z}∕2\mathbb{Z})}^3)\simeq {\mathrm{GL}}_3({𝔽}_2)$ is a homomorphism.

Such a homomorphism is characterized by the image $\phi (1)$ of $1\in \mathbb{Z}∕7\mathbb{Z}$, of order $7$. So $\phi {(1)}^7$ corresponds to a matrix $A\in {\mathrm{GL}}_3({𝔽}^2)$ which satisfies $A^7=I$. The identity $I$ has order $1$, and there are $48$ elements of order $7$ in ${\mathrm{GL}}_3({𝔽}^2)$ (see the sagemath computation below), one of them being

$$A=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right).$$

     sage: liste= []
     sage: G = GL(3, GF(2))
     sage: G.order()
     168
     sage: I = G(1);
     sage: for A in G:
     ....:     if A^7 == I:
     ....:         liste.append(A)
     ....:
     sage: liste
     [
     [1 0 0]  [0 1 0]  [0 0 1]  [1 1 1]  [1 1 0]             [1 1 1]
     [0 1 0]  [0 0 1]  [1 0 0]  [0 1 1]  [1 0 1]    (...)    [0 1 1]
     [0 0 1], [1 1 0], [0 1 1], [1 1 0], [0 1 0],            [1 0 1]
     ]
     sage: len(l)
     49

Thus there are exactly $49$ homomorphisms $\phi :\mathbb{Z}∕7\mathbb{Z}\to \mathrm{Aut}({(\mathbb{Z}∕2\mathbb{Z})}^3)\simeq {\mathrm{GL}}_3({𝔽}_2)$.

If $\phi$ is the trivial homomorphism corresponding to $I$, then $G\simeq S\times H$, so $H⊴G$, which is in contradiction with the hypothesis. Each of the $48$ matrices of order $2$ gives a non trivial homomorphism, and the group $G\simeq {(\mathbb{Z}∕2\mathbb{Z})}^3{⋊}_{\phi }\mathbb{Z}∕7\mathbb{Z}$ is non abelian.

In particular, the group $G_A$, whose law is given, for $\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right),\left(\begin{array}{c}\hfill a^{\prime }\hfill \\ \hfill b^{\prime }\hfill \\ \hfill c^{\prime }\hfill \end{array}\right)\in {𝔽}_2^3$ and $x,x^{\prime }\in \mathbb{Z}∕7Z$ by

$$\left(\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right),x\right)\left(\left(\begin{array}{c}\hfill a^{\prime }\hfill \\ \hfill b^{\prime }\hfill \\ \hfill c^{\prime }\hfill \end{array}\right),x^{\prime }\right)=\left(\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right)+{\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right)}^x\left(\begin{array}{c}\hfill a^{\prime }\hfill \\ \hfill b^{\prime }\hfill \\ \hfill c^{\prime }\hfill \end{array}\right),x{x}^{\prime }\right)$$

is a group of order $56$ such that the Sylow $7$-Sylow subgroup is not normal. Therefore this group is not isomorphic to the previous groups.

If $A$ is replaced by another matrix $B\in {\mathrm{GL}}_3({𝔽}_2)$ of order $7$, we show that $G_A\simeq G_B$. The corresponding homomorphisms ${\phi }_A$ and ${\phi }_B$ are given by

$$\begin{array}{lll}\hfill {\phi }_A\{\begin{array}{ccc}\hfill \mathbb{Z}∕7\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}({𝔽}_2^3)\hfill \\ \hfill \bar{x}\hfill & \hfill \mapsto \hfill & {\phi }_A(\bar{x})\{\begin{array}{ccc}\hfill {𝔽}_2^3\hfill & \hfill \to \hfill & {𝔽}_2^3\hfill \\ \hfill \left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right)\hfill & \hfill \mapsto \hfill & A^x\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right)\hfill \end{array}\hfill \end{array}& & \hfill \\ \hfill {\phi }_B\{\begin{array}{ccc}\hfill \mathbb{Z}∕7\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}({𝔽}_2^3)\hfill \\ \hfill \bar{x}\hfill & \hfill \mapsto \hfill & {\phi }_B(\bar{x})\{\begin{array}{ccc}\hfill {𝔽}_2^3\hfill & \hfill \to \hfill & {𝔽}_2^3\hfill \\ \hfill \left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right)\hfill & \hfill \mapsto \hfill & B^x\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right)\hfill \end{array}\hfill \end{array}& & \hfill \end{array}$$

(Since $A^7=B^7=I$, the maps ${\phi }_A(\bar{x})$ and ${\phi }_A(\bar{x})$ are well defined and are automorphisms of ${𝔽}_2^3$.)

The $48$ elements of order $7$ in $G$ are in two conjugacy classes of ${\mathrm{GL}}_3({𝔽}_2)$, the classes of

$$A=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\text{and}{A}_1=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right).$$

     sage: G = GL(3, GF(2))
     sage: A = matrix([[0,1,0], [0,0,1], [1,1,0]])
     sage: A1 = matrix([[0,0,1], [1,0,0], [0,1,1]])
     sage: A = G(A); A1 = G(A1); A, A1
     
     (
     [0 1 0]  [0 0 1]
     [0 0 1]  [1 0 0]
     [1 1 0], [0 1 1]
     )
     sage: A.conjugacy_class().list()
     
     [
     [0 1 0]  [0 1 1]  [0 1 0]  [1 0 1]  [0 1 1]  [1 1 1]  [1 1 1]  [1 1 0]
     [0 0 1]  [0 0 1]  [1 0 1]  [1 1 1]  [1 1 0]  [1 0 1]  [1 0 0]  [1 1 1]
     [1 1 0], [1 0 0], [1 0 0], [1 1 0], [1 1 1], [0 1 1], [1 0 1], [0 1 0],
     
     [0 1 0]  [0 0 1]  [0 0 1]  [1 0 1]  [0 0 1]  [1 1 0]  [1 1 1]  [0 1 0]
     [1 1 1]  [0 1 1]  [1 0 1]  [0 0 1]  [1 0 0]  [0 0 1]  [1 1 0]  [0 1 1]
     [0 1 1], [1 1 1], [0 1 0], [1 1 1], [1 1 0], [1 0 1], [1 0 0], [1 0 1],
     
     [1 1 0]  [0 1 1]  [1 0 1]  [1 1 1]  [0 0 1]  [0 1 1]  [1 1 0]  [1 0 1]
     [0 1 1]  [1 0 0]  [1 1 0]  [0 1 1]  [1 1 0]  [1 1 1]  [1 0 1]  [1 0 0]
     [1 0 0], [0 1 0], [0 1 0], [1 1 0], [0 1 1], [1 0 1], [1 1 1], [0 1 1]
     ]
     sage: A1.conjugacy_class().list()
     
     [
     [0 0 1]  [1 1 1]  [1 0 1]  [0 0 1]  [1 0 1]  [0 1 0]  [1 0 1]  [0 1 1]
     [1 0 0]  [1 1 0]  [1 0 0]  [1 1 0]  [1 1 1]  [1 0 1]  [1 1 0]  [1 0 0]
     [0 1 1], [0 1 1], [0 1 0], [0 1 0], [0 1 1], [0 1 1], [1 1 1], [1 0 1],
     
     [1 0 1]  [1 1 0]  [1 1 0]  [0 1 1]  [1 1 1]  [0 0 1]  [0 1 0]  [0 1 1]
     [0 0 1]  [1 0 1]  [0 0 1]  [1 1 0]  [1 0 0]  [0 1 1]  [0 1 1]  [1 1 1]
     [1 1 0], [0 1 0], [1 0 0], [1 0 0], [1 1 0], [1 1 0], [1 0 0], [0 1 0],
     
     [0 1 0]  [0 0 1]  [1 1 0]  [0 1 1]  [1 1 0]  [1 1 1]  [0 1 0]  [1 1 1]
     [0 0 1]  [1 0 1]  [1 1 1]  [0 0 1]  [0 1 1]  [1 0 1]  [1 1 1]  [0 1 1]
     [1 0 1], [1 1 1], [1 0 1], [1 1 1], [1 1 1], [1 0 0], [1 1 0], [1 0 1]
     ]
                                                                  

                                                                  
     

Note that if $Q=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$, then

$$A_1^3=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right){\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)}^{-1}=\mathit{𝑄𝐴}{Q}^{-1},$$

so

$$A_1^3=\mathit{𝑄𝐴}{Q}^{-1}.$$

Let ${𝒞}_A$ and ${𝒞}_{A_1}$ denote the conjugacy classes of $A$ and $A_1$.

• If $B\in {𝒞}_A$, then $B=\mathit{𝑃𝐴}{P}^{-1}$ for some $P\in {\mathrm{GL}}_3(𝔽2)$.

• If $B\in {𝒞}_{A_1}$, then $B=R{A}_1{R}^{-1}$ for some $R\in {\mathrm{GL}}_3(𝔽2)$. Therefore

  $$B^3=R{A}_1^3{R}^{-1}=R(\mathit{𝑄𝐴}{Q}^{-1})R^{-1}=(\mathit{𝑅𝑄})A{(\mathit{𝑅𝑄})}^{-1}=\mathit{𝑃𝑄}{P}^{-1}\text{where }P=\mathit{𝑅𝑄}.$$

In both cases there is some integer $k$ ( $k=1$ or $k=3$) and some $P\in {\mathrm{GL}}_3({𝔽}_2)$ such that

$$\mathit{𝑃𝐴}{P}^{-1}=B^k.$$

If $\sigma \in \mathrm{Aut}({𝔽}_2^3)$ is defined by $\sigma (X)=\mathit{𝑃𝑋}$, then for all $X\in {𝔽}_2^3$ (where $X^{\prime }=\mathit{𝑃𝑋}=\sigma (X)$), and all $\bar{x}\in \mathbb{Z}∕7\mathbb{Z}$,

$$\begin{array}{llll}\hfill ({\phi }_B^k(\bar{x})\circ \sigma )(X)& ={\phi }_B^k(\bar{x})(X^{\prime })& \hfill & \\ \hfill & =B^{\mathit{𝑘𝑥}}{X}^{\prime }& \hfill & \\ \hfill & =P{A}^x{P}^{-1}{X}^{\prime }& \hfill & \\ \hfill & =P{A}^{x}X& \hfill & \\ \hfill & =(\sigma \circ {\phi }_A(\bar{x}))(X).& \hfill & \end{array}$$

Therefore ${\phi }_B^k(\bar{x})\circ \sigma =\sigma \circ {\phi }_A(\bar{x})$ for all $\bar{x}\in K=\mathbb{Z}∕7\mathbb{Z}$, thus

$${\phi }_B^k(K)=\sigma {\phi }_A(K){\sigma }^{-1},$$

where $K=\mathbb{Z}∕7\mathbb{Z}$ is cyclic.

By Exercise 6,

$${(\mathbb{Z}∕2\mathbb{Z})}^3{⋊}_{{\phi }_A}\mathbb{Z}∕7\mathbb{Z}\simeq {(\mathbb{Z}∕2\mathbb{Z})}^3{⋊}_{{\phi }_B}\mathbb{Z}∕7\mathbb{Z}.$$

There is a unique group of order $56$ with a nonnormal Sylow $7$-subgroup, up to isomorphism.