TITLE: Alternative way to show that a simple group of order $60$ can not have a cyclic subgroup of order $6$
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: Here is another answer based on a simpler observation than my last, inspired by considering a comment by @pGroups on an omission. It is in the same spirit, but is distinct.
Note that a simple group of order $60$ must have six subgroups of order $5$ permuted transitively by conjugation (Sylow). There can't be one subgroup of order $5$ because it would be normal. This action on the subgroups of order $5$ gives an injective homomorphism into $S_6$.
Now note that the elements of order $6$ in $S_6$ are all odd permutations, so if $G$ has an element of order $6$ its image contains odd permutations. The even permutations in the image form a normal subgroup of index $2$, so $G$ cannot be simple.