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Is there a characterization of finite groups having no dihedral subgroup of order $2p$ for all odd primes $p$ dividing the order of the group?

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    $\begingroup$ A non-solvable example is the binary icosahedral group, of order 120: indeed its only element of order 2 is central. $\endgroup$ – YCor Mar 23 '13 at 14:05
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Yes, these are just the finite groups in which all involutions (elements of order $2$) lie in $O_{2}(G),$the largest normal $2$-subgroup of $G,$ using the Baer-Suzuki theorem. Clearly, if every involution of $G$ lies in $O_{2}(G)$, then all dihedral subgroups of $G$ are $2$-groups. On the other hand if $G$ has no dihedral subgroup of order $2p$ for any odd prime $p,$ then whenever $t$ is an involution of $G,$we see that for each $g \in G,$ the dihedral group $\langle t,t^{g} \rangle$ is a $2$-group, so $t \in O_{2}(G)$ by the Baer-Suzuki theorem.

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