Some notes before the question:
1- there are many questions in MSE asking about elements generating $S_n$ but they all involve more than one transposition to generate $S_n$ for example "$S_n$ is generated by elements of the form $(1k)$ [more than one element because $k$ varies] or "... generated by $ \{(1,2), (1,2,3,...,n) \} $".
2- By cyclic or generated by, considering the same meaning of generating a group by a single element and so called the cyclic group.
My question is : for which $i$ and $i$, $(i \ \ j)$ generated $S_n$, i.e. $S_n = \langle (i \ \ j) \rangle$ i.e. $S_n = {\{ (i \ \ j)^a | \text{fixed} \ i, j \ \text{and varying}\ a \ \in {\{1, \dots n}\} \ }\}$? And if the answer is no single transposition can generated $S_n$, is it possible with two transpositions and if so for what numbers i, j, k, m $S_n = \langle (i \ \ j), (k \ \ m) \rangle$ ?
I would appreciate any simple clear detailed explanation.
As it was said in the comments (by Captain Lama and Santiago Canez), this is not possible if $n ≥ 3$ because the transpositions $(1 \; 2)$ and $(2 \; 3)$ do not commute. So $S_n$ is not abelian, and therefore it is not cyclic (i.e. it can't be generated by $1$ element). You only have $S_2 = \langle (1 \; 2) \rangle$.
For $n ≥ 5$, $S_n$ is not generated by $2$ transpositions: let $(a \; b)$ and $(c \; d)$ two of them, and let $x \in \{1,...,n\}$ with $x \not \in \{a,b,c,d\}$. Then $(a \; x)$ is not in the subgroup $G$ generated by $(a \; b)$ and $(c \; d)$, because every element of $G$ fixes $x$.
However, for $n=3$, you have $S_3 = \langle (1 \; 2), (1 \; 3) \rangle$.