Alternative way to show that a simple group of order $60$ can not have a cyclic subgroup of order $6$ 
Suppose $G$ is a simple group of order $60$, show that $G$ can not have a subgroup isomorphic to $ \frac {\bf Z}{6 \bf Z}$.

Of course, one way to do this is to note that only simple group of order $60$ is $A_5$. So if $G$ has a cyclic subgroup of order $6$ then it must have a element $\sigma$ of order $6$, i.e. in (disjoint) cycle decomposition of $\sigma$ there must be a $3$ cycle and at least $2$ transposition, which is impossible in $A_5$.Hence,we are done.
I'm interested in solving this question without using the fact that $ G \cong A_5$. Here is what I tried:
Suppose $G$ has a subgroup say $H$ isomorphic to $ \frac {\bf Z}{6 \bf Z}$, then consider the natural transitive action $G \times \frac {G}{H} \to \frac {G}{H}$, which gives a homomorphism $\phi \colon G \to S_{10}$. Can some one help me to prove that $\ker \phi$ is nontrivial ?
Is there any other way to solve this question? Any hints/ideas?  

Starting out with a patchwork of Sylow theory. Surely something simpler is out there :-(
A group $G$ of order $60$ has either $1,4$ or $10$ Sylow $3$-subgroups. If it has only $1$, then that is normal. If it has only $4$, then we have a non-trivial homomorphism $f$ from $G$ to $S_4$ meaning that $\operatorname{ker}(f)$ is a non-trivial normal subgroup of $G$. 
So we can assume that there are ten distinct Sylow $3$-subgroups. Let $P$ be one of them. Its normalizer $K=N_G(P)$ has order six. Because $H$ certainly normalizes a Sylow $3$-subgroup w.l.o.g. $H=K$. So any element $x\in G$ of order six generates the normalizer of $\langle x^2\rangle$. Therefore the conjugates of $H$ have intersections of order at most two.
In other words there are $20$ elements of order three and $20$ elements of order six in $G$ - two in each conjugate of $H$. There is no room left for the necessary six Sylow $5$-subgroups that $G$ should have in order to be simple.
