# Showing $\Bbb{Z}_q \rtimes Q_8$ has the presentation $\langle x,y,z \mid x^q=y^4=z^4=[x,y]=1, y^z=y^{-1}, y^2=z^2, x^z=x^{-1} \rangle.$

Let $$G \cong \Bbb{Z}_q \rtimes Q_8$$. Then $$G$$ has a presentation as follows

$$\langle x,y,z \mid x^q=y^4=z^4=[x,y]=1, y^z=y^{-1}, y^2=z^2, x^z=x^{-1} \rangle.$$

I dont understand why $$x^z=x^{-1}$$?

(The action is conjugation)

• Please specify which homomorphism $Q_8 \mapsto \text{Aut}(\mathbb{Z}_q)$ is used to define the semidirect product $\mathbb{Z}_q \rtimes Q_8;$ until you add that to your question, the semidirect product is not well-defined and your question is unanswerable. Also, once you've specified that homomorphism, it would help if you also added to your question which elements of $\mathbb{Z}_q \rtimes Q_8$ correspond to the generators $x,y,z$. – Lee Mosher May 11 '17 at 12:28
• The action is conjugation in any semidirect product. To define the group you need to specify a homomorphism from $Q_8$ to the automorphsim group of $C_q$ (the cyclic group of order $q$). – Derek Holt May 11 '17 at 12:34
• There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$. – Derek Holt May 11 '17 at 12:36
• @Derek Holt Ok thats right $y^4=1$. I have a group $G \cong C_q \rtimes Q_8$. My teacher said its presentation is as above, I dont understand why $x^z=x^{-1}$? I know $x^z \in C_q$ – Hana May 11 '17 at 12:46
• I am not sure you understand what an abstract grounp presentation is. In your case, there are three generators, $x,y,z$. They generate a free group. Now we require certain relations to hold between the generators. For example $y^4=1$. This gives rise to an equivalence relation of the free group. Similarly, for the other conditions. So $x^z := zxz^{-1} =x^{-1}$ must also hold. When all these relations are factored in, the claim is that the resulting group $G$ is a semi-direct product. – Somos May 11 '17 at 23:00