Find a nonabelian subgroup of order $10$ in $D_{15}$.
I know that I have to show that it has a reflection. How could I prove that if the subgroup contains all rotations, it wouldn't be a subgroup?
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$\begingroup$ There is something odd about the phrasing of this question. A subgroup is never empty, and in this case it even has to have a specified order, so mentioning non-empty seems strange. $\endgroup$– Tobias KildetoftJan 29, 2013 at 20:55
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1$\begingroup$ @BabakSorouh The notation varies. Sometimes, the index is the order, sometimes half the order (the degree as a permutation group in the natural way). $\endgroup$– Tobias KildetoftJan 29, 2013 at 21:00
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1$\begingroup$ @BabakSorouh in my text it translates to having order 30 $\endgroup$– user5208Jan 29, 2013 at 21:01
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1$\begingroup$ @NickKidman Groups don't have idempotent elements. $\endgroup$– Tobias KildetoftJan 29, 2013 at 21:10
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1$\begingroup$ I can't believe people haven't agreed on $D_n$ instead of $D_{2n}$. We don't call the symmetric groups $S_1$, $S_2$, $S_6$, $S_{24}$, etc. /rant $\endgroup$– Paul OrlandJan 29, 2013 at 21:14
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3 Answers
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$D_{15}$ is the symmetries of a regular 15-gon. "Extend" a regular pentagon in Euclidean plane to a regular 15-gon so that any symmetry of the pentagon is in $D_{15}$. So, $D_{15}$ contains a copy of $D_5$.
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We are working on $$D_{2\times 15}=\langle \alpha,\beta\mid\alpha^{15}=\beta^2=(\alpha\beta)^2=1\rangle$$ We can also consider $\alpha,\beta$ as follows:
$$\alpha=(1,2,...15),~\beta=\begin{pmatrix} 1 & 2 & 3 & 4& ~...~14& 15\\ 1 & 15 & 14 & 13& ~...~3& 2\\ \end{pmatrix}$$
Clearly the only power of $\alpha$ of order $5$ is $\alpha^3$. Now set $H=\langle\alpha^3\rangle$. Using the relation of the group we can show that $H$ is a normal subgroup. NOe we take $H\langle\beta\rangle$.
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$\begingroup$ If you mean why upvotes aren't increasing my rep: once you've earned 200 points from upvotes, you cap for the day (the upvotes still appear as an additional vote, but you can earn at most 200 points from upvotes).You can earn unlimited points per day from "accepts". The new day starts in roughly 8 hours. When rep for the day new day returns to "zero" $\endgroup$– amWhyJan 30, 2013 at 15:57
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$\begingroup$ In roughly 8 hours (a little less than that now), the new day starts for everyone at stackexchange.com $\endgroup$– amWhyJan 30, 2013 at 16:33
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$D_{15} = \left<a,b | a^{15}=1, b^2=1, bab=a^{-1} \right>$
Consider the subgroup generated by $a^3$ and $b$.
answered Jan 29, 2013 at 21:17