Is this true? Let $G\neq A_5$ be a finite simple nonabelian group. Then $G$ has a cyclic subgroup of order $2p$ and a subgroup isomorphic to the dihedral group of order $2p$, for some prime $p$.
3 Answers
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 $A_5$.
According to the subgroup lattice http://homepages.ulb.ac.be/~tconnor/atlaslat/m11.pdf, the Mathieu group $M_{11}$ doesn't.
(But it does have a cyclic subgroup of order $2p$ and a dihedral subgroup of order $2q$, for some primes $p$ and $q$.)

$\begingroup$ Thanks for your answer. Is the Mathieu group $M_{11}$ the only counterexample? $\endgroup$ May 26, 2013 at 5:25

$\begingroup$ Wouldn´t think so, it was the first that I checked at the website homepages.ulb.ac.be/~tconnor/atlaslat I´m sure you´ll find more examples in there. $\endgroup$– IsaMay 26, 2013 at 5:36
If we only consider the case that the prime $p$ in the question is odd (which amounts to considering simple groups which do not have elementary Abelian Sylow $2$subgroups), another class of simple groups is $G = {\rm PSL}(2,q)$ when $q$ is a Mersenne prime greater than $3$. For any odd prime divisor $p \neq q$ of $G$, we see that $G$ has a cyclic Sylow $p$subgroup $P$ with $N_{G}(P)$ dihedral of order $2P$, so that $G$ has a dihedral subgroup of order $2p$ but no element of order $2p$ (note that $P$ itself need not have order $p).$ The only other odd prime divisor of $G$ is $q$ itself, and a Sylow $q$subgroup $Q$ of $G$ has $N_{G}(Q)$ of order $\frac{q(q1)}{2}$, which is odd. It follows that no nonidentity element of $Q$ is conjugate to its inverse in $G$ ( for if it were, the conjugation could be effected within $N_{G}(Q)$, which is clearly impossible). Thus $G$ has no dihedral subgroup of order $2q$.