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?  

Suppose $a$ has order $6$ and generates the subgroup $A$ of order $6$.
If $A$ is the only subgroup of order $6$ it is normal, so there must be more than one such subgroup. How many are there? Well the normaliser $N$ of $A$ contains $A$ as a subgroup and is not the whole of $G$. That means that it has order $12$ or $30$ (or, as comment suggests, 6 - see below for this).
A subgroup of order $30$ would have index $2$ and would therefore be normal. So the order must be $12$ and there must be $5$ subgroups of order $6$ which are conjugate to $A$. The action of $G$ by conjugation on these subgroups must be faithful (otherwise the non-trivial kernel would be a normal subgroup) and this gives an injective homomorphism of $G$ into $S_5$.
Now an element of order $6$ in $S_5$ is an odd permutation, so the image of $G$ in $S_5$ contains odd permutations. The even permutations in the image form a normal subgroup of index $2$. So if $G$ has a cyclic subgroup of order $6$ it cannot be simple.

Note the key element here that an element of order $6$ in $S_5$ is an odd permutation. The rest of the argument would apply to the nonabelian group of order $6$ as a subgroup, which has to map into $A_5$ if $G$ is to be simple.

For normaliser of order $6$, there are $10$ subgroups of order $6$ containing $20$ distinct elements of order $6$
Sylow gives us six subgroups of order $5$ (we can't have one - it would be normal) containing $24$ elements of order $5$
We can't have one subgroup of order $3$, nor can we have four (action by conjugation gives homomorphism to S_4 which would have a non-trivial kernel) so ten is the only possibility with $20$ elements of order $3$.
And that takes us over $60$ elements already, so the situation is impossible.
