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I'm trying to prove the dihedral groups are solvable for any Dn. I use the normal subgroup of all rotations, since the quotient of Dn/{rotations} is isomorphic to Z2 so it's abelian as well, so we get a solvable group.

Joey
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1 Answers1

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I think your idea is correct: There is an exact sequence $$ 1 \to \mathbb{Z}/n\mathbb{Z} \to D_n \to \mathbb{Z}/2\mathbb{Z} \to 1, $$ where the map $\mathbb{Z}/n\mathbb{Z} \to D_n$ identifies $\mathbb{Z}/n\mathbb{Z}$ with the rotations and $D_n \to \mathbb{Z}/2\mathbb{Z}$ is the quotient map. Since $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$ are solvable, so is $D_n$.

Artem Mavrin
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  • Thank you! but what does this sequence means? since I have only learnt one definition of the solvable group which is using a sequence of quotient groups and every quotient group is abelian. – Joey Dec 18 '15 at 21:54
  • @Mark another way of thinking about the "exact sequence" I mentioned is the following result: If a group $G$ has a normal subgroup $H$ such that $H$ and $G/H$ are both solvable, then $G$ is solvable. In this case $G=D_n$ and $H$ is the cyclic subgroup of rotations. – Artem Mavrin Dec 18 '15 at 22:20
  • Are you using the theorem that for a normal subgroup N of G, if N and G/N are solvable then G is solvable. and {rotations} is normal in $D_2n$ and {rotations} isomorphic to Zn is solvable since it's abelian, and $D_2n$/{rotations} is isomorphic to $Z_2$ is solvable since it's abelian as well, so we know that by the theorem $D_2n$ is solvable. – Joey Dec 18 '15 at 22:27
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    @Mark yes, that's the result I used. – Artem Mavrin Dec 18 '15 at 22:29