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Suppose we're given a Dihedral group $D_{30}$.
a) Find a cyclic subgroup H of order 10 in $D_{30}$. List all generators of H.
b) Let k and n be an integer such that k >= 3 and k divides n. Prove that $D_n$ contains exactly one cyclic subgroup of order k.

My attempt:
a) in $D_{30}$ we know that |r| = 30. So we can find a cyclic subgroup H generated by r such that it is of order 10. take < $r^{30/10}$ > = < $r^3$ >. Then < $r^3$ > contains the identity element e and powers of $r^3$, to $r^{27}$. Thus the generator of H is then $r^3$. Would this be okay?

b) Since k divides n, we can write n as n = kp for some p. Then we see that gcd(n,k) = k and thus:
|< r >| = |r| = n.

Then by Fundamental theorem of cyclic groups I can say that the group has exactly one subgroup of order k, ie: = since n = kp. And we are done.

This is my first course in Group Theory so I'm rather shaky and insecure about my proofs. Your comments and help would be greatly appreciated.

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notice that you didn't use the fact that $k\geq 3$ in b. Try to find where you need it (and then correct your proof) –  Prometheus Feb 2 '11 at 9:39
    
There are two conventions for the dihedral group. In one convention, $D_n$ is the group associated to the regular $n$-gon and has $2n$ elements. In the other, $D_{2n}$ is the dihedral group with $2n$ elements. It is always good practice to specify which one you are using (clearly, the former here). –  Arturo Magidin Feb 2 '11 at 16:08
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2 Answers

up vote 2 down vote accepted

You are using $D_{30}$ to represent the dihedral group of order $60$, with $r$ corresponding to the "rotation", and some other element, call it $s$, the "reflection"; that is, $$D_{30}=\left\langle r,s\Bigm| r^{30} = s^2 = 1,\quad sr = r^{-1}s\right\rangle.$$

Part (a): Your first part begins well enough, but it seems like you did not read the question carefully. The last part of part (a) asks you to find all generators of the subgroup $H$ you found.

You took $H=\langle r^3\rangle$. That's fine; this is a group of order $10$. But $r^3$ is not the only generator of this group. Perhaps you know that, in general, the cyclic group of order $n$, $C_n$ has $\varphi(n)$ generators, where $\varphi$ is Euler's function that counts the number of positive integers less than $n$ and relatively prime to $n$ (if you don't, then try to prove it). So $H$, being cyclic of order $10$, should have $\varphi(10) = 4$ generators. You've found one, there are three more to go.

Your part (b) also suffers a bit. You know that there is one and only one subgroup of $\langle r\rangle$ that has order $k$; that's fine. But you did not prove that every subgroup of $D_n$ of order $k$ must be a subgroup of $\langle r\rangle$! You need to show that too if you want your argument to hold. So you would need to show that the only elements of order $k$ all lie in $\langle r\rangle$; that's where you'll need the fact that $k\geq 3$. Perhaps you can show that every other element has other ideas about what their order is?

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For the first part, one solution is to use brute force as follow.

Let us consider $D_{30}$ acting on a set of cardinality $30$, generated by the two elements $(1, 2, 3, 4, 5,$ $6, 7, 8, 9, 10,$ $11, 12, 13, 14, 15,$ $16, 17, 18, 19, 20,$ $21, 22, 23, 24, 25,$ $26, 27, 28, 29, 30)$ and $(1, 30)(2, 29)(3,28)$ $(4, 27)(5, 26)(6, 25)$ $(7, 24)(8, 23)(9, 22)$ $(10, 21)(11, 20)(12, 19)$ $(13, 18)(14, 17)(15, 16)$.

Then the cyclic subgroup of order 10 of $D_{30}$ is generated by $(1, 28, 25, 22, 19,$ $16, 13, 10, 7, 4)$ $(2, 29, 26, 23, 20,$ $17, 14, 11, 8, 5)$ $(3, 30, 27, 24, 21,$ $18, 15, 12, 9, 6)$ and $(1, 25, 19, 13, 7)$ $(2, 26, 20, 14, 8)$ $(3, 27, 21, 15, 9)$ $(4, 28, 22, 16, 10)$ $(5, 29, 23, 17, 11)$ $(6, 30, 24, 18, 12)$.

This can be checked easily with Magma or GAP. However I agree it lacks subtelty. One can probably use part $b)$ to build a smarter answer.

For the second part, I would suggest this.

It is well known that a dihedral group can be generated by two involutions as follows: $$D_{n}=\langle s,t \mid s^2=t^2=e, (st)^n=e \rangle$$ On the other hand, a cyclic group of order $k$ is generated by a unique element of order $k$: $$C_k= \langle r \mid r^k=e \rangle$$ Now take an integer $k\geq 3$ such that $k\mid n$. Hence there exists $p \in \mathbb{N}$ such that $pk=n$. Considering $s,t$ as in the presentation of $D_n$, one has $$(st)^n = e$$ and thus $$((st)^p)^k = e$$ that is to say, there exists an element $r := (st)^p$ of order $k$ in $D_n$.

Therefore there exists $H < D_n$ such that $$H= \langle r \mid r^k=e \rangle$$ i.e. $H$ is a cyclic group of order $k$. Unicity is a consequence of the fact that $r$ is uniquely determined as soon as $k$ is not $2$. In this case, any involution generates a cyclic group of order $2$.

I hope I answered your question.

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It's not true that $r$ is uniquely determined, because various powers of $r$ will also have order $k$. Also, the uniqueness depends on the fact that all of the elements of $D_n$ that are not powers of $st$ have order 2. –  Derek Holt Feb 2 '11 at 14:59
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