# A presentation for $\mathbb Q_8\rtimes_{\phi}\mathbb Z_3$?

We know one of the presentation of $\mathbb Q_8$ is: $$\mathbb Q_8=\langle a,b,c|ab=c,bc=a,ca=b\rangle$$ and if we want to construct the semi-direct product of $\mathbb Q_8\rtimes\mathbb Z_3$; this can be carried out by defining a proper homomorphism, say $\phi$: $$\phi:=\mathbb Z_3\longrightarrow Aut(\mathbb Q_8)\cong\mathbb S_4$$ Usually, the groups which I had to examine, have been both cyclic, but this time one of them is the quaternion group, $\mathbb Q_8$. What I have learnt is to define a suitable homomorphism sending generators of groups to each other. So, here I should consider $\phi$ to send $x$ of order 3, as $\mathbb Z_3=\langle x\rangle$ to a correspondent element in $Aut(\mathbb Q_8)$.

My problem is to define a suitable homomorphism $\phi$ and then demonstrate an associated presentation of $\mathbb Q_8\rtimes_{\phi}\mathbb Z_3$. Thanks for the time you share.

• $a$ goes to $b$ goes to $c$ goes to $a$.
– user641
Sep 14, 2012 at 20:18
• @SteveD: You mean that I take $y\in Aut(Q_8)$ as (a,b,c) in S_4? Sep 14, 2012 at 20:29
• @SteveD: If so; then I should find an element of right hand side group which has the same order always. Right? Sep 14, 2012 at 20:31
• No. Order dividing, maybe. For example, you can take $Q_8\rtimes C_4$ where the generator of $C_4$ goes to $(a,b)$.
– user641
Sep 14, 2012 at 22:07
• @SteveD: Thanks. Sep 15, 2012 at 0:41

Let $$G=Q_8\rtimes \mathbb{Z}_3$$, and suppose that action of $$\mathbb{Z}_3$$ on $$Q_8$$ (by conjugation) is non-trivial. Since $$Q_8$$ has three subgroups of order 4: $$\langle i\rangle$$, $$\langle j\rangle$$, $$\langle k\rangle$$; the conjugation action of $$\mathbb{Z}_3$$ on $$Q_8$$ will permute these subgroups. As $$|\mathbb{Z}_3|=3$$, orbit of a subgroup will have order $$1$$ or $$3$$; hence if one subgroup is fixed, then all subgroups will be fixed by $$\mathbb{Z}_3$$.

If one (hence all) subgroups fixed, then consider action of $$\mathbb{Z}_3$$ on $$\langle i\rangle=\{1,-1,i,-i\}$$ by conjugation. Since $$-1$$ is unique element of order 2 here, it will be fixed by $$\mathbb{Z}_3$$. Hence, $$\mathbb{Z}_3$$ will permute $$\{i,-i\}$$ by conjugation. But, again, orbit of $$i$$ should have order $$1$$ or $$3$$; the only possibility is that orbit should be singleton. We conclude that, if $$\mathbb{Z}_3$$ fixes $$\langle i\rangle$$, then it fixes this subgroups pointwise, and similarly, it will fix $$\langle j\rangle$$, $$\langle k\rangle$$ pointwise. Therefore, action of $$\mathbb{Z}_3$$ on $$Q_8$$ is trivial, a contradiction.

Hence, the non-trivial action of $$\mathbb{Z}_3$$ must permute the subgroups $$\langle i\rangle$$, $$\langle j\rangle$$, $$\langle k\rangle$$ cyclically.

Now, we can easily define a homomorphism you wanted: if $$\mathbb{Z}_3=\langle z|z^3=1\rangle$$, define

$$z\mapsto \{ i\mapsto j, j\mapsto k, k\mapsto i\}$$, i.e. $$z^{-1}.i.z=j, z^{-1}.j.z=k$$, $$z^{-1}.k.z=i$$.

(Remark: this shows that there is only one non-trivial semidirect product of $$Q_8$$ by $$\mathbb{Z}_3$$; hence there is unique group $$G$$ such that $$G=Q_8 \rtimes_{1} \mathbb{Z}_3$$. The only such group is $$SL(2,\mathbb{Z}_3)$$.)

• The uniqueness follows from the more general fact that split cyclic extensions are classified by conjugacy classes in the automorphism group. In this case, since all subgroups of order 3 in $S_4$ are conjugate, there is only non-trivial semidirect product up to isomorphism.
– user641
Sep 15, 2012 at 6:58
• @Steve: yes; that's correct; but I simply determinded "action of $C_3$ on $Q_8$", without determination of $Aut(Q_8)$. Sep 15, 2012 at 7:42
• Thanks for you detailed answer. Indeed, taking $\alpha=\left(\begin{array}{cc} 1 & 1 \\1 & 2 \end{array}\right)$ and $\beta=\left(\begin{array}{cc} 2 & 2 \\1 & 0 \end{array}\right)$, the generators of $SL(2,3)$; the group that @Urban presented is constructed. Thanks for your attempt here. :) Sep 15, 2012 at 11:34

Pick any element of order $3$ in $S_4$ to get a homomorphism $\varphi \colon ~ \mathbb{Z}_3 \to \mathrm{Aut}(\mathbb{Q}_8)$. The presentation of the semi-direct product is then given by $$\mathbb{Q}_8 \rtimes_\varphi \mathbb{Z}_3 = \langle a,b,c, x \mid ab = c, bc = a, ca = b, x^3 = 1, a^x = a^{\varphi(x)}, b^x = b^{\varphi(x)}, c^x = c^{\varphi(x)} \rangle,$$ a disjoint union of presentations of the original groups plus the conjugation relations induced by the action $\varphi$.