For prime $p$, a transitive subgroup of $S_p$ containing a transposition is all of $S_p$

Because the action of $G$ on $\{0,1,\dots ,p-1\}$ is transitive, the orbit of any point has size $p$. Hence [L1] gives $p\mid \mathrm{\mid }G\mathrm{\mid }$. By [L2], the group $G$ contains an element $c$ of order $p$.

By [L3], a permutation of order $p$ in $S_p$ must be a $p$-cycle: every nontrivial cycle length divides $p$, so each is $1$ or $p$, and there must be one nontrivial cycle. Conjugating inside $S_p$, we may relabel so that $c=(01\dots p-1)\mathrm{.}$

Write the given transposition as $\tau =(ij)$ with $i\mathrm{\ne }j$, and put $d\equiv j-i(\mathrm{m}\mathrm{o}\mathrm{d}p)$, choosing $d\in \{1,\dots ,p-1\}$. For each $k$, $c^k\tau c^{-k}=(i+kj+k),$ so $G$ contains every transposition of the form $(xx+d)$, with indices read modulo $p$. Because $p$ is prime, $d$ is invertible modulo $p$, so the sequence $0,d,2d,\dots ,(p-1)d$ lists all $p$ symbols exactly once modulo $p$. Therefore the $p-1$ transpositions joining consecutive terms in that order all lie in $G$.

Let $\pi$ be the relabelling permutation carrying $k$ to $kd$ modulo $p$. By step 2.1 and [L5], the conjugates $\pi (kk+1){\pi }^{-1}$ for $0\le k\le p-2$ are exactly the $p-1$ transpositions joining consecutive terms in the ordering of step 2.1, so they lie in $G$. Since [L4] says the standard adjacent transpositions generate $S_p$, their conjugates also generate $S_p$. Hence $G$ contains a generating set of $S_p$, so $G=S_p$.