2
$\begingroup$

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 $p$.

$\endgroup$
1
  • 2
    $\begingroup$ 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. $\endgroup$ May 26, 2013 at 5:06

3 Answers 3

9
$\begingroup$

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 $A_5$.

$\endgroup$
3
$\begingroup$

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 $q$.)

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Isa
    May 26, 2013 at 5:36
3
$\begingroup$

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 $2q$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.