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Prove that every subgroup of $D_n$ , either every member of subgroup is a rotation or exactly half of them are rotations.

Intuitively, if every member is a rotation then they will form a subgroup because we can rotate them as much as we like (closure) and other properties will also be satisfied. But how do we prove that exactly half of them will be rotations form a subgroup. Please give Hints to start!

Thanks

GodotMisogi
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Gathdi
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  • You're finding that every subgroup is exactly either 100% or 50% rotations, not that a subset with exactly either 100% or 50% rotations must be a subgroup. As for proving it, the 100% is easy and done already. Now assume at least one element not a rotation and show that this requires half the elements altogether to not be rotations. – Nij May 21 '16 at 10:33
  • @Nij can you please elaborate? – Gathdi May 21 '16 at 10:45
  • For the first part: "subgroup therefore n% elements are x" is not the same statement as "n% elements are x therefore subgroup". For the second part: suppose there is at least 1 non-rotation element in the subgroup. What does that imply about other elements of it? – Nij May 21 '16 at 10:50
  • @Nij I was looking at table for $D_4$. If we compose one relection with other rotations we get other reflections. That means if half are rotations and we take one reflection and compose them we get another half as reflections. But this is not a proof – Gathdi May 21 '16 at 10:57
  • Try proving it by using $$D_n=<\sigma, \rho | \sigma^2=\rho^2=e, \sigma \rho^k \sigma= \rho^{-k}>$$ – M. Van May 21 '16 at 12:36
  • If $H \subset D_n$ is a subgroup and it contains only rotations you are done, then suppose it contains some $\rho^k \sigma$, try to look at what else the subgroup must contain in this case, and what it does not... – M. Van May 21 '16 at 12:58
  • By the way, this is also a special case of the theorem on index $2$ subgroups. The rotations form an index $2$ subgroup of $D_n$, and what you are required to prove is that any subgroup is either contained entirely inside the subgroup of rotations, or is half inside and half outside it. – M. Vinay May 21 '16 at 18:44
  • @Gathdi why not? That's just what other commenters are suggesting - you simply need to give it a formalism that extends to all $n$. – Nij May 21 '16 at 18:54

1 Answers1

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This is what the question is really asking you:

Let $D_n$ be the dihedral group of order $2n$ and $R$ the order $n$ group of rotations.

The if $H$ is any subgroup of $D_n$, either $[H:H \cap R] = 1$ or $[H:H \cap R] = 2$.

If $H \subseteq R$, then $H \cap R = H$, and we have $[H:H \cap R] = [H:H] = 1$.

Otherwise, we can invoke (via the second isomorphism theorem):

$[H: H \cap R] = [HR:R]$.

Now $HR$ is a subgroup of $D_n$ properly containing $R$, and $[D_n:R] = 2$, so $HR = D_n$. Can you continue?

David Wheeler
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  • i do not know about these notations and terms. can you provide simple answer to where i am stuck – Gathdi May 23 '16 at 01:34
  • @Gathdi Maybe if you explain which notations/terms that you do not understand it can be explained to you? – Irregular User May 23 '16 at 01:42
  • @irregular isomorphism, [h : h intersection r ] – Gathdi May 23 '16 at 02:37
  • The notation $[G:H]$ means the index of a subgroup $H$ in $G$, for a finite group (as we have here) this is just $|G|/|H|$ (and if $H$ is normal is equal to $|G/H|$), which is just the number of (right or left) cosets of $H$ in $G$. If you do not yet know the second isomorphism theorem, you can use the product set counting formula $|HK| = \dfrac{|H|\cdot|K|}{|H \cap K|}$ to establish $\dfrac{|HR|}{|R|} = \dfrac{|H|}{|H\cap R|}$ – David Wheeler May 23 '16 at 02:39