If $|G|=120$ and $|H|=24$ and $H$ has at least two Sylow $2$-subgroups, then does $G$ acts faithfully on $G/H$?
I know that, if $n$ is the number of Sylow $2$-subgroups of $H$, then by Sylow's theorems, $n\mid \frac{24}{8}=3$, so $n=3$ because $n\geq 2$.
Also, I know that the kernel of the action of $G$ on $G/H$ by left translations is $\bigcap\limits_{g\in G}gHg^{-1}$.
Actually I am trying to do the last item from the exercise 10 of section 1.13 of Jacobson's Basic Algebra 1 about a characterization of $S_5$, but I do not know if the information I obtained about $H$ is sufficient to it.
I tried to follow the hints from someone in this page: An Abstract Characterization of $S_5$ using involutions and their centralizers, that suggests me to prove first that the kernel $K$ of the action is a proper subgroup of $H$, that $K$ is centralized by an element of order $5$, looking at $\mathrm{Aut}(K)$, so it cannot contain elements of order $2$, and to prove that $K=Z_3$ would be normal in $G$, leading to a contradiction.
But I do not know even how to prove that $K\neq H$. I tried to do this: If $K=H$, then $\bigcap\limits_{g\in G}gH^{-1}g=H$, so $H$ is a normal subgroup of $G$, then $G/H$ is a group isomorphic to $Z_5$ and the action would be a homomorphism from $G$ to $G/H\cong Z_5$, but after this I am stuck.
The answer is 'maybe'. The answer is 'yes' for the group $G=S_5$, with $H=S_4$. But for the group $G=S_4\times C_5$ with $H=S_4$ the answer is 'no'. Since $H\lhd G$ in the second case, of course $G$ does not act faithfully.
For the group $(A_4\times C_5).2$, where the $2$ acts diagonally to make $S_4$ and $D_{10}$, the answer is 'no', but the kernel $K$ is the subgroup $V_4$ of $S_4$.