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.
Best Answer
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.