Is this true?
Let $G\neq A\_5$ be a finite simple non-abelian group. Then $G$ has a cyclic subgroup
of order $2p$ and a subgroup isomorphic to the dihedral group of order $2p$, for some prime $p$. 



a question about finite simple non-abelian groups
For $f>1$ let $G = {\rm SL}\_2({\bf F}\_{2^f})$
(a.k.a. $L\_{\phantom.2}(2^f)$ in ATLAS notation). Then $G$ is simple
and each element has exponent either $2$ or a factor of $2^f \pm 1$.
Hence $G$ has no cyclic subgroup of order $2p$ for any prime $p$
(not even $2$). For $f=2$ we recover the example of $A\_5$.


