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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?

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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).

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  • $\begingroup$ Actually I got stuck in a problem of topology where I need some idea aboubt group theory, So if you don't mind, can you tell me is that fact correct or not if I replace dihedral group by the group of symmetries of a regular polytope? $\endgroup$ – Anubhav Mukherjee Sep 9 '16 at 18:19
  • $\begingroup$ If you don't mind can I edit this question with the last comment? $\endgroup$ – Anubhav Mukherjee Sep 9 '16 at 18:20
  • $\begingroup$ I've posted this as a separate qes here math.stackexchange.com/questions/1920723/… $\endgroup$ – Anubhav Mukherjee Sep 9 '16 at 18:38
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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 $4$.

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  • $\begingroup$ This is a nice observation. Thanks. $\endgroup$ – Anubhav Mukherjee Sep 9 '16 at 18:13
  • $\begingroup$ Actually I got stuck in a problem of topology where I need some idea aboubt group theory, So if you don't mind, can you tell me is that fact correct or not if I replace dihedral group by the group of symmetries of a regular polytope? $\endgroup$ – Anubhav Mukherjee Sep 9 '16 at 18:27
  • $\begingroup$ I've posted it as a separate qes here math.stackexchange.com/questions/1920723/… $\endgroup$ – Anubhav Mukherjee Sep 9 '16 at 18:38
  • $\begingroup$ Also of order eight, but abelian, $(\mathbb{Z}/2\mathbb{Z})\times (\mathbb{Z}/2\mathbb{Z})\times (\mathbb{Z}/2\mathbb{Z})$ can be seen to be a counterexample. $\endgroup$ – Jeppe Stig Nielsen Sep 10 '16 at 11:49

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