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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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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. –  Tobias Kildetoft Jan 29 '13 at 20:55
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@BabakSorouh The notation varies. Sometimes, the index is the order, sometimes half the order (the degree as a permutation group in the natural way). –  Tobias Kildetoft Jan 29 '13 at 21:00
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@BabakSorouh in my text it translates to having order 30 –  user5208 Jan 29 '13 at 21:01
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@NickKidman Groups don't have idempotent elements. –  Tobias Kildetoft Jan 29 '13 at 21:10
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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 –  orlandpm Jan 29 '13 at 21:14

3 Answers 3

up vote 3 down vote accepted

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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Nice approach to the problem! +1 –  amWhy Jan 30 '13 at 0:47
    
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" –  amWhy Jan 30 '13 at 15:57
    
Wait for the new day! ;-) –  amWhy Jan 30 '13 at 16:13
    
In roughly 8 hours (a little less than that now), the new day starts for everyone at stackexchange.com –  amWhy Jan 30 '13 at 16:33

$D_{15}$ 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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$D_{15} = \left<a,b | a^{15}=1, b^2=1, bab=a^{-1} \right>$

Consider the subgroup generated by $a^3$ and $b$.

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