All non abelian groups of order $56$, when $\mathbb Z_7\triangleleft G$

I am calculating all non abelian groups of order $56$ when $\mathbb Z_7$ is a normal subgroup ( Given in the exercise $7$ of §5.5, Dummit Foote). I have searched for related posts on this site but couldn't able to find something that discusses on this particular topic. It would be great if someone provide me link on this site discussing on this topic . However, here is my progress.

Let $S$ be the Sylow $2$ subgroup and throughout the discussion I will call $\mathbb Z_7=\left<d\right>$, where $d=1$. My calculations are as follows :

$\boxed{\text{case }(1) : S=\mathbb{Z_2}\times\mathbb{Z_2}\times\mathbb{Z_2}=\left<a,b,c\right>}$

Since all elements of $\mathbb{Z_2}\times\mathbb{Z_2}\times\mathbb{Z_2}=\left<a,b,c\right>$ has order $2$, so for any homomorphism $\phi\colon\left<a,b,c\right>\to Aut(\mathbb Z_7)$, each of $\phi(a),\phi(b),\phi(c)$ has order either $1$ or $2$.

The element of $Aut(\mathbb Z_7)$ having order $1$ is the identity automorphism $\alpha_1\colon1\mapsto1$ and that having order $2$ is the automorphism $\alpha_2\colon1\mapsto6$ because $1\mapsto6\mapsto6\cdot6=36=1$ in $\mathbb Z_7$. So the possible homomorphisms are : $\begin{array}{c|c}\phi_1&\phi_2\\\hline a\mapsto\alpha_1 & a\mapsto \alpha_1\\b\mapsto\alpha_1& b\mapsto\alpha_2\\c\mapsto \alpha_2&c\mapsto\alpha_2\end{array}\tag*{}$

In case of the homomorphism $\phi_1$,

$a\cdot d=d, b\cdot d=d, c\cdot d=d^{-1}$,

where the dot represents the action on $d$.

$\therefore ada^{-1}=d,bdb^{-1}=d,cdc^{-1}=d^{-1}\text{ with }a^2=b^2=c^2=d^7=1$

Since $a$ and $b$ centralize both $c$ and $d$, so in this case the group must be $\begin{align}G_{\phi_1}&=\left<a|a^2=1\right>\times\left<b|b^2=1\right>\times\left<c, d|c^2=d^7=1,cdc^{-1}=d^{-1}\right>\\&=\mathbb Z_2\times\mathbb Z_2\times D_{14}\\&=\mathbb Z_2\times D_{28}\end{align}$

In case of the homomorphism $\phi_2$, $a\cdot d=d, b\cdot d=d^{-1}, c\cdot d=d^{-1}$. $\therefore ada^{-1}=d,bdb^{-1}=d^{-1},cdc^{-1}=d^{-1}\text{ with }a^2=b^2=c^2=d^7=1$.

But the group $G_{\phi_2}$ formed by these relations is exactly same as the above group : if we invert all the elements of $G_{\phi_1}$, we get $G_{\phi_2}$. So $$G_{\phi_1}\cong G_{\phi_2}$$

$\boxed{\text{case }(2) : S=\mathbb{Z_4}\times\mathbb{Z_2}=\left<a,b\right>}$

By the same argument in case $1$, the possible homomorphisms $\mathbb Z_4\times\mathbb Z_2\to Aut(\mathbb Z_7)$ are :

$\begin{array}{c|c|c}\psi_1&\psi_2&\psi_3\\\hline a\mapsto\alpha_1& a\mapsto\alpha_2&a\mapsto \alpha_2\\b\mapsto\alpha_2& b\mapsto\alpha_2&b\mapsto \alpha_1\end{array}\tag*{}$

In case of the homomorphism $\psi_1$,

$G_{\psi_1}=\left<a,b,d|a^4=b^2=d^7=1, ab=ba ,ada^{-1}=d, bdb^{-1}=d^{-1}\right>$

Hence it is easily seen to be $$G_{\psi_1}\cong\mathbb Z_4\times D_{14}$$

In case of the homomorphism $\psi_2$,

$G_{\psi_2}=\left<a,b,d|a^4=b^2=d^7=1,ab=ba ,ada^{-1}=d^{-1},bdb^{-1}=d^{-1}\right>$

Now, $(ab) \cdot (d)=(ab)d(ab)^{-1}=a(bdb^{-1})a^{-1}=ad^{-1}a^{-1}=d$.

And order of $ab$ is $4$ and $ab$ centralises both $b$ and $d$.

So the relations determined by $\psi_2$ are same as the relations determined by $\psi_1$.

So,

$G_{\psi_2}=\left<ab\right>\times\left<b,d\right>=\mathbb Z_4\times D_{14}\cong G_{\psi_1}$

And the homomorphism $\psi_3$ produces the group $\begin{align}G_{\psi_3}&=\left<a,b,d|a^4=b^2=d^7=1, ab=ba ,ada^{-1}=d^{-1}, bdb^{-1}=d\right>\\&=\left<b|b^2=1\right>\times\left<a,d|a^4=d^7=1,ada^{-1}=d^{-1}\right>\\&=\mathbb Z_2\times\left<a,d|a^4=d^7=1,ada^{-1}=d^{-1}\right>\end{align}$

$\boxed{\text{case }(3) : S=\mathbb{Z_8}=\left<a\right>}$

The only possible homomorphism $\pi\colon\mathbb Z_8\to Aut(\mathbb Z_7)$ that gives rise to a non abelian group is : $\pi\colon a\mapsto\alpha_2$.

Here the group is $G_\pi=\mathbb Z_7\rtimes_{\pi}\mathbb Z_8$.

$\boxed{\text{case }(4) : S=\mathbb Q_8=\left<i ,j\right>}$

Since here $S$ is non abelian, we need to take the trivial homomorphism $S\to Aut(\mathbb Z_7)$ into account.

The possible homomorphisms are :

$\begin{array}{c|c|c|c}\theta_1&\theta_2&\theta_3&\theta_4(\text{ trivial })\\\hline i\mapsto\alpha_1& i\mapsto\alpha_2&i\mapsto \alpha_2&i\mapsto\alpha_1\\j\mapsto\alpha_2& j\mapsto\alpha_1&j\mapsto \alpha_2&j\mapsto\alpha_1\end{array}\tag*{}$

In case of the homomorphism $\theta_1$, the relations between the elements are : $idi^{-1}=d, jdj^{-1}=d^{-1}, i^2=j^2=-1, d^7=1, ij=-ji$

I don't know the name of this group $G_{\theta_1}$. I would like to know if there is some name or explicit notation for this group.

The group $G_{\theta_2}$ formed by the homomorphism $\theta_2$ is isomorphic to $G_{\theta_1}$ because we are just interchanging the roles of $i$ and $j$.

In case of the homomorphism $\theta_3$, we have the relations : $idi^{-1}=d^{-1}, jdj^{-1}=d^{-1}, i^2=j^2=-1, d^7=1, ij=-ji$.

Now in the relation $j\cdot d=jdj^{-1}= d^{-1}$ , if we hit by $i$, $(ij)d(ij)^{-1}=id^{-1}i^{-1}=d$.

So writing $ij=k$ the relations determined by $\theta_3$ are $idi^{-1}=d^{-1}, kdk^{-1}=d$, which is exactly the same as the group determined by $\theta_1$ or $\theta_2$.

So $G_{\theta_1}\cong G_{\theta_2}\cong G_{\theta_3}$.

The trivial homomorphism $\theta_4\colon\mathbb Q_8\to Aut(\mathbb Z_7)$ corresponds to $G_{\theta_4}=\mathbb Q_8\times\mathbb Z_7$

$\boxed{\text{case }(5) : S=D_8=\left<r ,s\right>}$

Here the possible homomorphisms are :

$\begin{array}{c|c|c|c}\eta_1&\eta_2&\eta_3&\eta_4(\text{ trivial })\\\hline r\mapsto\alpha_1& r\mapsto\alpha_2&r\mapsto \alpha_2&r\mapsto\alpha_1\\s\mapsto\alpha_2& s\mapsto\alpha_1&s\mapsto \alpha_2&s\mapsto\alpha_1\end{array}\tag*{}$

The corresponding groups are

as alex mentioned in the comment :

$\begin{align}G_{\eta_1}&=\{r, s,d|r^4=s^2=d^7=1,srs=r^{-1},rdr^{-1}=d,sds=d^{-1}\}\\&=\{r,s,d|(rd) ^{28}=s^2=1,s(rd)\cdot s(rd)=s(dr)\cdot s(rd)=sd(rsr)d=sd(s)d=(sds)d=d^{-1}d=1\}\\&=D_{56}\end{align}$

$G_{\eta_2}=\{r, s,d|r^4=s^2=d^7=1,srs=r^{-1},rdr^{-1}=d^{-1},sds=d\}$

and $G_{\eta_3}=\{r, s, d|r^4=s^2=d^7=1,srs=r^{-1},rdr^{-1}=d^{-1},sds=d^{-1}\}$

In $G_{\eta_2}$, if we hit the relation $s\cdot d=sds^{-1}=d$ by $r$, we get,

$(rs)\cdot d=(rs)d(rs)^{-1}=r(sds^{-1})r^{-1}=rdr^{-1}=d^{-1}$

So $G_{\eta_2}=\{r, s,d|r^4=s^2=d^7=1,srs=r^{-1},rdr^{-1}=d^{-1},(rs)d(rs)^{-1}=d^{-1}\}\cong G_{\eta_3}$

And here too I come up with the group $G_{\eta_2}$ for the first time. So it would be great if someone please help me to understand these groups.

The trivial homomorphism $\eta_4\colon D_8\to Aut(\mathbb Z_7)$ gives rise to the group $G_{\eta_4}$ which is the direct product $D_8\times\mathbb Z_7$

Most importantly, I would like to know whether there are any flaws in my calculations. Thank you.

In a comment citing a reference, I said there were 10 nonabelian groups of order 56, which is true.

But I forgot that you are looking for nonabelian groups that have a normal Sylow-7 subgroup. Consider:

• $\newcommand{\Aut}{\operatorname{Aut}}\Aut(\mathbb{Z}_8)\cong \mathbb{Z}_4$, so there is no nontrivial mapping $\mathbb{Z}_7\to\Aut(\mathbb{Z}_8)$.
• $\Aut(\mathbb{Z}_4\times\mathbb{Z}_2)$ has order not divisible by $7$, so there is no nontrivial mapping $\mathbb{Z}_7\to\Aut(\mathbb{Z}_4\times\mathbb{Z}_2)$.
• $\Aut(\mathbb{Z}_2^3)\cong \mathbb{Z}_4$ has order $7\cdot3\cdot1$. OK, so here there is a nontrivial mapping $\mathbb{Z}_7\to\Aut(\mathbb{Z}_2^3)$. This group of order $21$ must have a unique Sylow-7 subgroup, so there is only $1$ such mapping (up to isomorphism), leading to one nonabelian group of order 56 where its Sylow-7 subgroup is not normal.

OK, so the above was to establish that you should be expecting to find $9$ nonabelian groups of order 56 having a normal Sylow-7 subgroup. Your arguments have found $8$.

You are missing one in your discussion of $\psi$ (Case 2). You claim $G_{\psi_1}\cong G_{\psi_2}$, but there are errors in this line of justification: $$(ab)d(ab)^{-1}=a(bdb^{-1})a^{-1}=ad^{-1}a=d$$

In fact, $G_{\psi_1}\not\cong G_{\psi_2}$, and that accounts for the missing group. (Also you did not investigate $G_{\psi_3}$, but that one turns out to be isomorphic to $G_{\psi_2}$.)

As noted in the comments, I felt that you skipped over what would be $\phi_3$, with both $a,b\mapsto\alpha_2$. (This is not consequential, because it leads to a $G_{\phi_3}$ that is isomorphic to $G_{\phi_1}$. It seemed out of step with the rest of the argument, if the goal was to convince the reader you were covering all the bases.)

Also as noted in the comments, I think much of the arguement could be condensed using shuffling of generators.