I am studying some exercises about semi-direct product and facing this solved one:
Show that the order of group $G=\langle a,b| a^6=1,a^3=b^2,aba=b\rangle$ is $12$.
Our aim is to show that $|G|\leq 12$ and then $G=\mathbb Z_3 \rtimes\mathbb Z_4=\langle x\rangle\rtimes\langle y\rangle$. So we need a homomorphism from $\mathbb Z_4$ to $\mathrm{Aut}(\mathbb Z_3)\cong\mathbb Z_2=\langle t\rangle$ to construct the semi-direct product as we wish: $\phi:=\begin{cases} 1\longrightarrow \mathrm{id},& \\\\y^2\longrightarrow \mathrm{id},& \\\\y\longrightarrow t,& \\\\y^3\longrightarrow t, \end{cases} $ Here, I do know how to construct $\mathbb Z_3 \rtimes_{\phi}\mathbb Z_4$ by using $\phi$ and according to the definition. My question starts from this point: The solution suddenly goes another way instead of doing $(a,b)(a',b')=(a\phi_b(a'),bb')$. It takes $\alpha=(x,y^2), \beta=(1,y)$ and note that these elements satisfy the relations in group $G$. All is right and understandable but how could I find such these later element $\alpha, \beta$?? Is really the formal way for this kind problems finding some generators like $\alpha, \beta$? Thanks for your help.