Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
$a$ 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? –  B. 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? –  B. 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. –  B. S. Sep 15 '12 at 0:41
show 2 more comments

2 Answers

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 $\varphi$.

share|improve this answer
add comment

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

share|improve this answer
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. :) –  B. S. Sep 15 '12 at 11:34
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.