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im trying to show that $D_6 \cong D_3\times \mathbb Z_2$ as groups.

$D_n$ is the is the dihedral of order 2n

I know that it is possible to form all of the points in $D_6$ and in $D_3$ as compositions of $R$ the rotation operation (such that every vertex move one to the left) and T the reflection on one of the symmetry axis, so i tried to write it done and joining point that act the same, but for no use.

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  • For you, $;D_n;$ is the dihedral of order $;n;$ or of order $;2n;$ ? – Timbuc Nov 05 '14 at 16:00
  • yes, sorry, i edited the question now. – sha Nov 05 '14 at 16:03
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    @Timbuc: Since there is no dihedral group of order $3,$ isn't that obvious? – Cameron Buie Nov 05 '14 at 16:08
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    @cameronbuie: considering that the question currently states (incorrectly, as per your comment) that $D_n$ is of order $n$, I'm not sure it's so obvious... – Semiclassical Nov 05 '14 at 16:14
  • @Semiclassical: Certainly not universally. It's possible that the OP isn't using the term "order" in the usual way, I suppose. – Cameron Buie Nov 05 '14 at 16:16
  • @CameronBuie, nothing is obvious in this site as many times deeply nonsensical question have been posted, as you must surely know. In this case, I didn't even pay attention to the right side. – Timbuc Nov 05 '14 at 16:28
  • @sha Anyway: the claim is false as $;D_6\cong S_3;$ and this is a non-abelian group, so it can't be the direct product of two non-abelian groups. – Timbuc Nov 05 '14 at 16:30
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    @Timbuc: If $D_6\cong S_3,$ then $D_3$ doesn't exist, so is neither abelian nor non-abelian. The claim is true under the correct interpretation of $D_n,$ and meaningless under the one stated by the OP. – Cameron Buie Nov 05 '14 at 17:03
  • @CameronBuie, I'm taking here, under the OP's definition, $;D_3\cong A_3\cong C_3;$= the cyclic group of order three. It never matters though, as whatever non-trivial $;D_3;$ could be in the OP's mind , $;S_3;$ is not a direct product of two non-trivial groups of order less than six. Nevertheless, and as noted before, nonsensical questions appear rather often here, so interpretations are open. – Timbuc Nov 05 '14 at 17:14

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There are exactly two possibilities, here, as far as I can see.

One possibility is that $D_n$ is intended (as you say in your question) to denote the dihedral group with $n$ elements (which is what is usually meant by saying a group has order $n$). However, all dihedral groups have an even number of elements, so under that interpretation, there is no such thing as $D_3,$ and so $D_3\times\Bbb Z_2$ doesn't exist! Since $D_6$ exists (under this interpretation, it is the group of symmetries of an equilateral triangle), it certainly isn't isomorphic to a non-existent group.

The other interpretation, which I suspect is what was intended is that $D_n$ is the dihedral group on $n$ vertices--the group of symmetries of a regular $n$-gon--which has $2n$ elements (order $2n$). Suspect this is the intended meaning since for any odd $n,$ we have that $D_{2n}\cong D_n\times\Bbb Z_2.$ For a proof of this more general result (and an idea of how to prove the specific case in question), see here.

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