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.

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.
