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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 $p$.

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Are you allowing $p=2$ (so the "dihedral group" of $2p$ elements is a 4-group)? If not then $A_6$ seems to be another example because it has no element of exponent $2p$ for $p$ odd. – Noam D. Elkies May 26 '13 at 5:06
up vote 7 down vote accepted

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 $A_5$.

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According to the subgroup lattice, 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 $q$.)

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Thanks for your answer. Is the Mathieu group $M_{11}$ the only counterexample? – majid arezoomand May 26 '13 at 5:25
Wouldn´t think so, it was the first that I checked at the website I´m sure you´ll find more examples in there. – Isa May 26 '13 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 $2|P|$, 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 $2q$.

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