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I just showed that $D_4$ is solvable since it is isomorphic to $\mathbb{Z}_2 x \mathbb{Z}_2$ so since $\mathbb{Z}_2$ is solvable so is $D_4$. Does something similar extend to $D_5$ or is there a better way of showing both?

John Smith
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  • It's not exactly relevant, but there appears to be a notation clash if $D_4$ refers to a group with $4$ elements; $D_5$ could only possibly refer to a group with $10$ elements. – pjs36 Nov 07 '17 at 04:24
  • @pjs36 D4 has 8 elements – John Smith Nov 07 '17 at 04:41
  • What is the meaning of $\mathbb{Z}_2 x \mathbb{Z}_2$ here? Direct product of two copies of $\mathbb{Z}_2$ ? But this would have only $4$ elements... – ladisch Nov 07 '17 at 12:58

1 Answers1

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All dihedral groups $D_n$ have an abelian normal subgroup order $n$ (in fact it is cyclic) with the corrresponding quotient abelian (of order 2, to be precise) and so they are solvable.

P Vanchinathan
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