Generators of the symmetric group $S_n$ [transpositions] It is clear that $S_n$ is generated by the adjacent transpositions, i.e., $S_n$ = $\langle (1,2),(2,3),...,(n-1,n) \rangle $.
So I took this idea by different way. Let me take for example $S_3$.
It is clear that $S_3 = \langle (1,2), (1,2,3)\rangle $
Then, $S_3 = \langle (1,2), (1,3)\rangle =  \langle (2,3), (1,2)\rangle = \langle (1,3), (2,3)\rangle$
Observing each case, $(1,2)$ and following transposition $(1,3)$ sharing common element "$1$". With the same method $(2,3)$ and following tranposition $(1,2)$ sharing the common element "$2$". Likewise $(1,3)$ and $(2,3)$ sharing the common element "$3$"
So I conducted induction that make my self the statement $(*)$
$(*)$ $S_n = \langle (i_1,i_2),(i_2,i_3),(i_3,i_4),...,(i_{n-1},i_n)\rangle $   for $i_k \in \{1,2,3,...,n\} $.
(Here if $k \neq l \Rightarrow i_k \neq i_l $.)
As you've known, the induction itself doesn't not guarantee true at all because it is just guess. So, does the statement $(*)$ always hold? If true would you suggest the proof idea? (If not, what is a counterexample?) Any answer or comment appreciated. thanks.
 A: Your $i_k$ implies a permutation $i = (j \mapsto i_j : j = 1, 2, \ldots, n)$, and you will find that each $i$-adjacent transposition $(i_k, i_{k+1})$ can be written $i(k, k+1)i^{-1}$, therefore your list of generators generate the same as $\langle i,(k,k+1) : k=1,2,\ldots,n-1\rangle$, which is clearly all of $S_n$, since you already observed just having the adjacent transpositions is enough.
To see why $\langle(i_k, i_{k+1}) : k = 1, 2, \ldots, n-1\rangle=\langle i,(k,k+1) : k=1,2,\ldots,n-1\rangle$, observe the following:
One direction of containment is easy (the RHS generates all of $S_n$ and so in particular it generates the LHS), for the other, note that $i$ itself, as a permutation, can be written as a sequence of transpositions $i = \tau_1\tau_2\cdots\tau_\ell$, and so
$$\prod_{j} i\tau_ji^{-1}=i\left(\prod_{j} \tau_j\right)i^{-1}= iii^{-1} = i$$
Thus the LHS can generate $i$ itself, and so it can also recover any transposition $(k, k+1)$
A: HINT
If it is $(1 \ i_k)$ and $(2 \ i_j)$  then 
$(1 \ 2)=(1\ i_j)...(i_a \ i_b)(i_k\ i_a)(2\ i_k)$
