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well i found this question answered in

Is the dihedral group $D_n$ nilpotent? solvable?

but the answer involves the sylow thm which i haven't studied. i'm studying bhattacharya and this question appears before the books starts to talk about the structure thms of groups...i wonder if it is possible to answer the question at this moment

thanks for your help

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Mathcho
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  • They are all solvable. Can you prove that? – Derek Holt Sep 02 '15 at 17:03
  • yes that is in fact easy...what bothers me is the nilpotency – Mathcho Sep 02 '15 at 21:26
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    Here is one possible approach in summary. If $n$ is a power of $2$ then $D_n$ is a $2$-group and hence nilpotent. If $n$ is odd then $D_n$ has trivial centre and so it cannot be nilpotent. Otherwise $D_n = 2^km$ with $m$ odd, and $D_n$ has a quotient group isomorphic to $D_m$, so it is not nilpotent. – Derek Holt Sep 02 '15 at 21:58
  • @DerekHolt what is a $2$-group?? I keep trying to look this up and figure out what it is, but I can't find anything even remotely relevant! –  Mar 15 '17 at 18:38
  • For positive integer $\alpha$ and prime $p$, a group of order $p^{\alpha}$ is a p-group. In regards to Sylows Theorem: for any group of order $p^{\alpha}m$ (where $(m,p)=1$, there always exisits at least one $p$-subgroup – Andrew Murphy Dec 06 '17 at 00:09
  • @DerekHolt Clearly $D_n$ is nilpotent when $n=2^k$, could you give some hints that how we can know the class of nilpotency. – MANI Apr 29 '19 at 12:20

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