How to prove that $\langle\{ (1,2),(1,2,3) \}\rangle=\mathfrak{S}_3$ 
Prove that
  $\{ (1,2),(1,2,3) \}$ Generating set of a symmetric group $(\mathfrak{S}_3,\circ )$

SOlution provided by book 
we 've
$(1,2,3)(1,2)(1,2,3)^2=(2,3)$ and $(1,2,3)^2(1,2)(1,2,3)=(1,3)$
since transposition Generating $(\mathfrak{S}_3,\circ )$ then $\{ (1,2),(1,2,3) \}$ Generating set of $(\mathfrak{S}_3,\circ )$


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*I didn't understand that solution would someone explain it to me and  how we can show such permutation is generator of a symmetric group and why they need to calculate  $(1,2,3)^2$ at left of $(1,2,3)(1,2)$.. 

 A: You have that $\mathfrak S_3=\{1, (12),(13),(23),(123),(132)\}$.
You wrote $(23)$ and $(13)$ as combinaison of $(12)$ and $(123)$. By the way, $(132)=(13)(32)$, and thus you can also express it as a combinaison of $(12)$ and $(123)$. Therefore, these two elements generate the group.
A: \begin{align}
  & \operatorname{Order}((1,2))=2\,\,\,,\,\,\operatorname{Order}((1,2,3))=3\,\, \\ 
 & \left\langle \left. (1,2),(1,2,3) \right\rangle  \right.=\left\{ {{(1,2)}^{i}}{{(1,2,3)}^{j}}|\,i=\,0\,,1\,\,\,\,,\,\,j=0,1,2\, \right\} \\ 
 & (1,2)=(1,2) \\ 
 & (1,2,3)=(1,2,3) \\ 
 & (1,2)(1,2,3)=(1,3) \\ 
 & (1,2,3)(1,2,3)=(1,3,2) \\ 
 & (1,2){{(1,2,3)}^{2}}=(1,2)(1,3,2)=(2,3) \\ 
 & {{(1,2,3)}^{3}}=e \\ 
\end{align}
A: The general idea is to reduce the problem to something you already know. 
If you know that transpositions generate the symmetric group, and if you can use your generators to get all the transpositions, then your generators must generate the symmetric group as well. 
The actual mechanics of expressing $(1, 3)$ as a product of (powers and inverses of) generators can be a messy, trial-and-error process. It's analogous to $\epsilon$-$\delta$ proofs from calculus: you generally do a bunch of background work to find a suitable $\delta$ for a given $\epsilon$. But this rarely shows up in the actual proof; you just show that the $\delta$ you found works, without explaining where you got the $\delta$ (at least, it's not necessary to explain this).
So it is with the products of generators that yield transpositions: if you can find some product, that's all you need (and, as has been pointed out, it need not be unique). It would be tedious and confusing for the author to list out the "failed" attempts (products of generators that aren't $(1, 3)$), so you just see the final result: something that works.
That's truly "why" one generator is squared, and why the product is arranged the way it is; it works, and that's all the proof requires!
A: $\left\langle \left.(1,2,3) \right\rangle  \right.$ has two left cosets
$$\left\langle \left.(1,2,3) \right\rangle  \right. = \left\{(1), (1,2,3), (1,3,2)\right\}$$
and
$$(1,2)\left\langle \left.(1,2,3) \right\rangle  \right.=\left\{(1,2), (1,3), (2,3) \right\}$$
in $\mathfrak S_3$. Therefore
$$\mathfrak S_3=\left\langle \left.(1,2,3) \right\rangle  \right.\cup (1,2)\left\langle \left.(1,2,3) \right\rangle  \right.\subseteq \left\langle \left. (1,2),(1,2,3) \right\rangle  \right. \subseteq \mathfrak S_3.$$
