# Non-trivial semidirect product $\mathbb Z_3\rtimes Q_8$ is isomorphic to dicyclic group of order $24$

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.

• 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 $G \to G/H$. – Rylee Lyman Nov 9 '19 at 0:28
• You could use GAP. – Shaun Nov 9 '19 at 1:43

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 $$(6,2,2)$$.