Let $G$ be a group acting on $X=\{a,b,c,d,e\}$. Given that there exist two elements $g,h\in G$ so that $$ g(a)=b,\ g(b)=c, \ g(c)=d, \ g(d)=e, \\ h(a)=b, \ h(b)=a, \ h(c)=d, \ h(d) =c. $$ Show that $G$ is not solvable.
My attempt: I looked at the homomorphism $\varphi:G\to\operatorname{Aut} X\cong S_5, g\mapsto(x\mapsto g(x))$. We can see that $g$ maps to $(12345)$, and $h$ maps to $(12)(34)$. Here is where I got stuck - how does this help showing $G$ is not solvable?
Let $H = \langle(1~2)(3~4), (1~2~3~4~5) \rangle \subseteq A_5$. Note that $H$ has elements of orders $2$ and $5$ so its order is a multiple of $10$. It's also easy to see that $H$ contains a $3$-cycle (the commutator of the generating elements is a $3$-cycle), so in fact $30$ must divide the order of $H$.
But $A_5$ is simple so it cannot have any subgroups of index $2$. Thus, $\vert H \vert \neq 30$ so $\vert H \vert = 60$ and $H=A_5$. We know that $A_5$ is not solvable and it is a homomorphic image of $G$, so $G$ has a composition series with a non-solvable factor, which means $G$ cannot be solvable.