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My question:

How can I prove isomorphism

$\mathbb Z_3\rtimes Q_8=\langle w,z\mid w^{12}=1, z^2=w^6, zwz^{-1}=w^{11} \rangle \cong\langle a,b,c\mid a^6=b^2=c^2=abc\rangle$?

Background:

The left side occurs in classifying group of order $24$, it is non-trivial simidirect product $\mathbb Z_3\rtimes Q_8$,

and the right side is dicyclic group of order 24.

Consider $Q_8=\langle y, z\mid y^4=1, z^2=y^2, zyz^{-1}=y^3 \rangle$ acting on $\mathbb Z_3=\langle x\rangle$ non-trivially.

We have homomorphism $\varphi:Q_8\to\operatorname{Aut}(\mathbb Z_3)\cong\mathbb Z_2$.

Subgroups of order $4$ in $Q_8$ are all isomorphic to $\mathbb Z_4$, so $\mathbb Z_3\rtimes Q_8$ is unique under isomorphism.

Suppose $\operatorname{ker}\varphi=\langle y \rangle$, then $\mathbb Z_3\rtimes Q_8$ has presentation

$\langle x,y,z\mid x^3=y^4=1, z^2=y^2, zyz^{-1}=y^3, yxy^{-1}=x, zxz^{-1}=x^2 \rangle$.

Let $x=w^4,y=w^3$, this can be reduced to $\langle w,z\mid w^{12}=1, z^2=w^6, zwz^{-1}=w^{11} \rangle$.

Groups of order $24$ and GAP show this group is isomorphic to dicyclic group of order 24,

i.e. $\langle w,z\mid w^{12}=1, z^2=w^6, zwz^{-1}=w^{11} \rangle \cong\langle a,b,c\mid a^6=b^2=c^2=abc\rangle$.

So how can I prove these two groups are isomorphic?

Thanks for your time and effort.

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    $\begingroup$ Just a note: we write $G = H\rtimes K$ when $H$ is a normal subgroup of $G$, $G/H \cong K$ and there is a section $K \to G$ of the map $G \to G/H$. $\endgroup$ Commented Nov 9, 2019 at 0:28
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    $\begingroup$ You could use GAP. $\endgroup$
    – Shaun
    Commented Nov 9, 2019 at 1:43

1 Answer 1

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According to groupprops about dicyclic group,

the left side is canonical presentation of dicyclic group of order $24$,

and the right side treats it as binary von Dyck group with parameters $(6,2,2)$.

Their equivalence is given here: Equivalence of presentations of dicyclic group.

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