a question about finite simple non-abelian groups 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: 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 $A_5$.
A: 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 $q$.)
A: 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 $2|P|$, 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 $\frac{q(q-1)}{2}$, which is odd. It follows that no non-identity 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 $2q$.
