Let $A_{34}$ be the alternating group of degree 34, $G_3$ be a Sylow $3$-subgroup of $A_{34}$ and $H=N_{A_{34}}(G_3)$. If $H$ is maximal subgroup of $A_{34}$ of index $34$, then $H\cong A_{33}$?
I want to prove that $A_{34}$ does not have exactly $34$ Sylow $3$-subgroups. If $A_{34}$ has $34$ Sylow $3$-subgroups, then $|G:N_G(G_3)|=34=2.17$, where $G=A_{34}$ and $G_3$ is a Sylow $3$-subgroup of $A_{34}$. For some reasons, we can prove that $H=N_G(G_3)$ is a maximal subgroup of $G=A_{34}$ of index $34$. Now if we can prove that $H=N_G(G_3)\cong A_{33}$, then we get a contradiction as $A_{33}$ is a simple group, hence $A_{34}$ does not have $34$ Sylow $3$-subgroups.
Yes. Consider the Schreier action of $A_{34}$ on $A_{34}/H$. It gives a homomorphism from $A_{34}$ to $S_{34}$. Since $A_{34}$ is simple the image is $A_{34}$ with the natural action on a $34$-element set. Now $H$ is the stabilizer of a point of that action. So it is isomorphic to $A_{33}$.