I was thinking that $$\alpha\beta=\beta_6$$ $$\beta\alpha=\beta_2$$ $$\beta^2\alpha= \beta_3$$
But I don't know what to do next or is this the right way of solving it,note I have only one example on this topic,,,if there is any link that may help me to learn this topic will be great.
Here $\alpha$ is the element $(123456)$ and $\beta$ is the element $(12)(36)(45)$. This is disjoint cycle notation of course. We need to prove that $\alpha$ has order $6$, that $\beta$ has order $2$, and that $\beta \alpha \beta = \alpha^5$.
The claims on order are obvious. $\alpha^6 = e$. $\beta^2 = e$.
Now to examine $\beta \alpha\beta$. We have the following permutations given by $\beta\alpha\beta$: $$1 \rightarrow 2 \rightarrow 3 \rightarrow 6$$ $$2 \rightarrow 1 \rightarrow 2 \rightarrow 1$$ $$3 \rightarrow 6 \rightarrow 1 \rightarrow 2$$ $$4 \rightarrow 5 \rightarrow 6 \rightarrow 3$$ $$5 \rightarrow 4 \rightarrow 5 \rightarrow 4$$ $$6 \rightarrow 3 \rightarrow 4 \rightarrow 5$$
That is to say $\beta\alpha\beta = (165432) = \alpha^5$.
So indeed, $G = \langle \alpha, \beta \rangle \cong D_{12} $.