generation of symmetric group of prime degree Let $p$ be a prime, $a=(1,2,...,p)$, $b$ is an odd permutation satisfying $b^{-1}ab\notin \langle a \rangle$ in $S_p$, I found that $\langle a,b \rangle$ is always the symmetric group $S_p$ when p is small ($p \leq 23$, use GAP and Magma), but I have no idea to prove it. Is it always true? Can you prove it or find a counterexample for it? Thanks for any help!

(Edited from the previous versions, which had many mistakes. The answer to the question seems to be yes.)
See "Jones, Gareth A. Cyclic regular subgroups of primitive permutation groups. J. Group Theory 5 (2002), no. 4, 403-407." for a classification of primitive subgroups of $S_n$ containing an $n$-cycle.
The groups $G$ that occur in the classification are the following:
(1.) $G = A_n$, $G = S_n$
(2.) $C_p \leq G \leq AGL_1(p)$
(3.) $G = PSL_2(11)$, $G = M_{11}$, $G = M_{23}$.
(4.) $PSL_d(q) \leq G \leq P \Gamma L_d(q)$ with $n = (q^d - 1)/(q-1)$.
To prove the result you claim, you should check that when $n$ is prime, in cases (3.) and (4.) you have $G \leq A_n$. For (3.) you have $G \leq A_n$ since $G$ is simple.
So the only interesting case is $G = P \Gamma L_d(q)$ in $S_n$ with $n = (q^d -1)/(q-1)$. We can assume that $PSL_d(q)$ is non-abelian simple, which holds if $d > 2$ or $q > 3$. Let's prove:

Lemma: $PGL_d(q) \leq A_n$ if and only if $d$ is odd or $q$ is even.
Proof: Let $\xi \in \mathbb{F}_q$ be a primitive element. We have $PGL_d(q) = \langle PSL_d(q), x \rangle$, where $x$ is the image of a diagonal matrix $diag(\xi,1,\ldots,1)$ in $PGL_d(q)$. Here $PSL_d(q) \leq A_n$ since it is simple, so we need to check when $x \in A_n$.
The cycle decomposition of $x$ consists of $(q^{d-1} -1)/(q-1) = 1 + q + \cdots + q^{d-2}$ cycles of length $q-1$. Hence $x \in A_n$ if and only if $d$ is odd or $q$ is even.

In the case where $n = (q^d -1)/(q-1)$ is prime we have $d$ odd, so in that case $PGL_d(q) \leq A_n$.
Then $P \Gamma L_d(q) = \langle PGL_d(q), \phi \rangle$, where $\phi$ is a field automorphism. The order of $\phi$ is $e$, where $q = p^e$ with $p$ prime. For $n = (p^{ed} -1)/(p^e - 1)$ to be prime $e$ must be odd. So $\phi$ has odd order, implying $\phi \in A_n$ and $P \Gamma L_d(q) \leq A_n$.
