Let $G=\langle a,b \rangle$ with $ord(a)=7, ord(b)=11$
$|X|=8$
Show that the group action $$\phi: G \times X \rightarrow X$$ $$(g,x) \mapsto g.x$$ is not transitive
I know that $\phi$ is transitive if and only if there is only one orbit.
I also know that the length of the orbit is equal to $|G/G_x|$, where $G_x$ is the stabilizer of $x$.
I know that the said stabilizer is a subgroup of $G$, and I have read that $|G/G_x|=|G| / |G_x|$, but would that not require $G_x$ to be normal?
Also, what is the order of $G$? Naively I would think it is $77$, but I am not sure whether this be deduced from the order of the two generators? From what I have read the answer is no. (The reference was the word problem for groups) Please correct me if I am wrong.
I am struggling with this task because I have no description of what $\phi$ does to $X$ and I am not sure of the order of $G$. If I had the order of $G$, I would have candidates for orders of $G_x$ and thus for the lengths of the orbits. (If the formula for factor groups applies for stabilizers).
Any hints and clarifications to the task are welcome.
My suggestion here would be to reinterpret the notion of group action. Instead of thinking about it as a map $\phi : G\times X\to X$, think of it as a map $G\to \newcommand\Sym{\operatorname{Sym}}\Sym(X)$. Then since the cardinality of $X$ is 8, we have that the map is $G\to S_8$, which means that $b\mapsto 1$, since the order of $b$ is $11$, which doesn't divide $8!$. Thus the image of $G$ in $S_8$ is either trivial or cyclic of order 7. Neither of which can act transitively.