How to prove a group fixes at least one point? Here is the two part question I am struggling with:
For the first part I have no idea at all. So if anyone could give me some ideas for that.
For the second bit I have the following which is not complete.
Suppose for a contradiction that $A_6$ acts transitively on a set of size $8$ then by the orbit stabiliser theorem we get: $8=\frac{|A_6|}{|\text{Stab}_G(x)|}$ now by the first part $A_5$ fixes at least one point so choose that point to be $x$ then this point is fixed by at least as many permutations of $A_6$ as it was of $A_5$ so $|\text{Stab}_G(x)| \geq |A_5|=120$ thus $8\leq\frac{|A_6|}{120}=8$ which is not a contradiction if we could prove that $A_6$ actually has more permutations that fix $x$ we would be done but I don't see it.
Can someone please help me on these two parts. I am not really sure about m argument as well it may be wrong.
Thanks!!

 A: Suppose that the action of $G$ on $X$, $|X|=8$ is transitive. Let $x\in X$, $|G|=8\times |G_x|$ where $G_x$ is the stabilizer of $x$. This is impossible, Since $8$ does not divide $60$. $X$ is the disjoint union of its orbits under $G$. There exists at least two orbits since the action of $G$ on $X$ is not transitive. There are not two orbits of size $\geq 5$, so you have at least an orbit of cardinal $<5$. The hypothesis implies that every element of this orbit is fixed by $G$.
A: By definition, $A_5$ acts on the orbits. The cardinal of an orbit, according to the hint, either is equal to $1$ or is at least $5$. On a set of cardinal $8$, this shows  at most one orbit can have cardinal $>1$. Furthermore the cardinal of an orbit is a divisor of $\lvert A_5\rvert=60$, less than $8$. Thus it is one of $5$ or $6$ (eliminating $2,3,4$ by the first argument). We conclude there are at least two fixed points.
As to the second part you have a miscalculation: $\;\dfrac{\lvert A_6\rvert}{120}=\dfrac{360}{120}=3,\;$ so you do have a contradiction.
