It is well-known that any finite group can be realised as a subgroup of a permutation group. But is it true that any finite group can be realised as a subgroup of a dihedral group?
2026-04-22 14:31:50.1776868310
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Finite groups as subgroups of dihedral groups
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No. Dietrich Burde's reason is quite nice, but there are easy elementary concrete counterexamples. Here's one:
$Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$ has six elements of order $4$, but no dihedral group has this property: all reflections in a dihedral group have order 2, and the only rotations of order $4$ are rotations by $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ so a dihedral group has at most two elements of order $4$.
No, because every dihedral group $D_n$ is solvable, and hence every subgroup of $D_n$ is solvable, too.
Actually, more can be said. Every subgroup of $D_n$ is either cyclic or dihedral, see Theorem $3.1$ of K. Conrad's notes. Therefore, say, $Q_8$ cannot be embedded (which we see by Doeke's answer).