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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.

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$a$ goes to $b$ goes to $c$ goes to $a$. – user641 Sep 14 '12 at 20:18
@SteveD: You mean that I take $y\in Aut(Q_8)$ as (a,b,c) in S_4? – Babak S. Sep 14 '12 at 20:29
@SteveD: If so; then I should find an element of right hand side group which has the same order always. Right? – Babak S. Sep 14 '12 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 '12 at 22:07
@SteveD: Thanks. – Babak S. Sep 15 '12 at 0:41
up vote 2 down vote accepted

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 $\varphi$.

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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 $SL(2,\mathbb{Z}_3)$.)

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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 '12 at 6:58
@Steve: yes; that's correct; but I simply determinded "action of $C_3$ on $Q_8$", without determination of $Aut(Q_8)$. – Marshal Kurosh Sep 15 '12 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. :) – Babak S. Sep 15 '12 at 11:34

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