Show that $\mathcal D_{12}\cong C_2\times \mathfrak S_3$

Question

Let $\mathcal D_{12}=\left<a,b\mid a^6=b^2=1,b^{-1}ab=a^{-1}\right>$, $\mathfrak S_3=\left<(12),(123)\right>$ and $C_2=\left<g\mid g^2=1\right>$. I would like to show that $\mathcal D_{12}\cong \mathfrak S_3\times C_2$. I set $c=(12)$ and $d=(123)$. Then, $$\mathfrak S_3=\{1,c,d,d^2,cd,cd^2\}$$ $$\mathcal D_{12}=\{1,a,a^2,a^3,a^4,a^5,ab,a^2b,a^3b,a^4b,a^5b\}$$ and $$C_2=\{1,g\}.$$ So, I can easily make association, like $$1\longmapsto (1,1)$$ $$a\longmapsto (cd,1)$$ $$a^2\longmapsto (d,1)$$ $$a^3\longmapsto (c,1)$$ etc... but is there an isiest way to do it ?