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.

Solutions Collecting From Web of "A presentation for $\mathbb Q_8\rtimes_{\phi}\mathbb Z_3$?"

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 $\varphi$.

Let $G=Q_8\rtimes \mathbb{Z}_3$, and suppose that action of $\mathbb{Z}_3$ on $Q_8$ (by conjugation) is non-trivial. Sicne $Q_8$ 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 easilt 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 $SL(2,\mathbb{Z}_3)$.)