Lemma on number of Sylow $p$-subgroups $P$ and index of $P$ in its normaliser correct? In this article I found the following lemma:

Let $p$ be prime and $G$ be a transitive subgroup of $S_p$, and suppose $|\textrm{Syl}_p(G)|>1$. Then $|N_G(P):P|>1$, where $P$ is a Sylow $p$-group and $N_G(P)$ is the normaliser of $P$ in $G$.

Why doesn't the following argument disprove it?
We can assume that $G$ is not the cyclic group with $p$ elements (otherwise $G=P$ and there is nothing to prove) but because $G$ is transitive its order is divisible by $p$. Therefore, the contraposition of the claim is: if $|N_G(P):P|=1$ then $|\textrm{Syl}_p(G)|=1$. But the condition means that $P$ is not normal because if it were normal then $N_G(P)=G\neq P$. But if $P$ is not normal then there must be more than 1 Sylow $p$-group and $|\textrm{Syl}_p(G)|>1$ as a Sylow $p$-group is unique if and only if it is normal.
The article is "An Elementary Proof of the Simplicity of the Mathieu Groups $M_{11}$ and $M_{23}$" by Robin Chapman, 1995.
 A: The only transitive group $G$ of prime degree $p$ with $N_G(P)=P$ for $P$ a Sylow $p$-subgroup is $G=P$ itself. So your proof doesn't contradict the lemma, it just broadens its applicability.
Here is the adjustment to the proof in the paper:

Proof: Suppose $G$ is a (faithful) transitive group on $p$ points and let $P$ be a Sylow $p$-subgroup of $G$. Let $N=N_G(P)$. Then $|G| = |P|[N:P][G:N]$. Note $|P|=p$. Set $m=[G:N]$ and $r=[N:P]$. Then $|G| = prm$. Since intersections of distinct conjugates of $P$ intersect in the identity, we get $m(p-1)$ elements of order $p$. Consider a point stabilizer $H$. It has index $p$, and so has no elements of order $p$. In other words, $|H|=rm$ and $H\cap P = 1$.
Suppose $r=1$. Then $m(p-1) = mrp-m = |G|-m$ elements have order $p$, and only $m$ elements have other orders. Thus there are only $m$ possible elements left in $G$ to make up $H$. Since $|H|=rm=m$, that means $H$ is exactly those $m$ elements. However, the same is true for all the point stabilizers, $H^g$, so that those exact same elements are used in every $H^g$. Said more simply, $H=H^g$ for all $g$ and $H$ is normal.
A faithful transitive group is exactly one in which $\bigcap\limits_{g\in G} H^g = 1$, but this group has $\bigcap\limits_{g\in G} H^g = \bigcap\limits_{g\in G} H = H$. We conclude that $H=1$ and the group is regular. Since $|H|=rm=1$ we get $m=1$ as well, and $|G|=prm=p$ so that $G=P$.☐

The paper uses slightly different language, but hopefully both are clear. Another way to finish is a classic exercise: somewhere in the middle we have shown $G=H \cup P$, but that can only happen if $G=H$ (no) or $G=P$, yes.
