Commutator subgroup of $G=\langle a,b,c,d:a^4=b^2=c^2=d^q=1,ab=ba,ac=ca,bc=cb,d^{-1}ad=b,d^{-1}bd=c,d^{-1}cd=ab \rangle$

Question

Let $$G=\langle a,b,c,d:a^4=b^2=c^2=d^q=1,ab=ba,ac=ca,bc=cb,d^{-1}ad=b,d^{-1}bd=c,d^{-1}cd=ab \rangle$$ be a group of order $8q$, where $q$ is an odd prime. I need to compute its commutator for some research. But I do not that how to proceed.Please help me to short out this problem.

Answer 1

Let $H$ be the subgroup of $G$ generated by $a$, $b$, and $c$. Observe that $H$ is isomorphic to $C_2^3$, where $C_n$ is the cyclic group of order $n$. Prove that $H$ is a normal subgroup of $G$. Now, $G/H$ is a group of order $q$, which must then be isomorphic to the cyclic group $C_q$ (i.e., $G/H$ is abelian).

If $K$ denotes the commutator subgroup of $G$, then we must have $K\leq H$ since $G/H$ is abelian. You now have to show that $K=H$.

We can determine $G$ precisely from the given information. Identify $H\cong C_2^3$ with the vector space $\mathbb{F}_2^3$, and identify $a$, $b$, and $c$ with the vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ respectively. Then, the conjugation by $d^{-1}$ (i.e., the map $x\mapsto d^{-1}xd$ for $x\in H$) is given by the matrix $$\mathbf{D}:=\begin{bmatrix}0&0&1\\1&0&1\\0&1&0\end{bmatrix}\,.$$ Thus, $\mathbf{D}^7$ is the identity matrix (over $\mathbb{F}_2$). However, $\mathbf{D}^q$ is also the identity matrix (as $d^q=1$). This shows that $q=7$. Consequently, $G$ is isomorphic to $\mathbb{F}_2^3\rtimes \langle \mathbf{D}\rangle\cong C_2^3\rtimes C_7$ with the group multiplication $$(\mathbf{x},\mathbf{D}^k)\cdot (\mathbf{y},\mathbf{D}^l):=(\mathbf{x}+\mathbf{D}^k\mathbf{y},\mathbf{D}^{k+l})$$ for all $\mathbf{x},\mathbf{y}\in\mathbb{F}_2^3$ and $k,l\in\mathbb{Z}/7\mathbb{Z}$. (The element $d$ of $G$ is identified with $(\mathbf{0},\mathbf{D}^{-1})$, for example.)

In general, if a nonabelian group $G$ of order $8q$, where $q$ is an odd prime natural number, has a normal subgroup isomorphic to $C_2^3$, then $q\in\{3,7\}$. This is because $G$ is a semidirect product of $C_2^3$ with $C_q$. We must then have a nontrivial group homomorphism $\tau:C_q\to \text{Aut}(C_2^3)$ (since $G$ is nonabelian). However, $\big|\text{Aut}(C_2^3)\big|=7\cdot 6\cdot 4$ and $\tau$ is injective. Therefore, $q\in\{3,7\}$. (For the OP's problem, the relations amongst the generators $a$, $b$, $c$, and $d$ eliminate the possibility that $q=3$.)

As a final note, if $q$ is an odd prime natural number not equal to $7$ and $G$ is the group with the given presentation, then from our earlier work, we can deduce that $d=1$. Thus, $d^{-1}ad=b$ implies $a=b$. Similarly, $b=c$ and $c=ab$. This shows that $a=b=c=1$. Therefore, $G$ is the trivial group.