Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $D_{n}$ is the Dihedral group of degree $n$ (order $2n$) where n is an odd number, can I then conclude that the numbers of elements of order 2 in $D_{n}$ is equal $n?$

I suppose there are the following types of elements in $D_{n}$: b, $a^{i}$, $a^{i}b$ for some $0\leq i < n$ .

share|cite|improve this question
yes, if the order of the group is odd then if one of the $a^i$ had order 2 then $a^{2i}=e\rightarrow 2i= n$ which is a contradiction as n is odd. – hmmmm Mar 31 '12 at 19:00
up vote 2 down vote accepted

$\bf{Convention:}$ Let $n$ be an odd number greater that or equal to $3$. Let $D_n$ denote the group of symmetries of regular $n-$ gon. Note that $|D_n|=2n.$

Yes, you're right. The elements of order $2$ in the group $D_{n}$ are precisely those $n$ reflections. Note that these elements are of the form $r^ks$ where $r$ is a rotation and $s$ is the reflection; $1 \leq k \leq n$.

In fact, we get some information about the Sylow Structure of $D_n$. Since, the number of elements of order $2$ in a group is the number of subgroups of order $2$, and that its exponent in $|D_n|$ is unity, we have that, the number of Sylow $2-$ subgroup of $D_n$ is $n$.

Now, the inquisitive$^\dagger$ reader would have observed that: For each $n$ such that $n \equiv 1 \bmod 2$, there is a group with exactly $n$ Sylow $2-$ subgroups.

Now, one can more generally ask:

Given a prime $p$, is there a group such that there are exactly $n$ Sylow $p-$ subgroups whenever $n \equiv 1 \bmod p$?

The even case of this problem is not hard to answer either: Note that the reflections are always of order $2$. So, for a regular $n-$ gon, with $n$ even, we have another element of order $2$-Rotation of the polygon by $\pi$. Note that these are the only elements of order $2$. This is done algebraically, via presentations, in Dylan's answer.

$\dagger$ It should go without saying (?) that I was not inquisitive and I realised I could ask this question only after reading one of Keith Conrad's blurbs. I cannot fish that link out now, but will add whenever I found that out. I am also reasonably sure, there are references where some examples for which the answer is negative has been discussed. My hearty thanks to him for leaving a commment here.

share|cite|improve this answer
Ah, I'm bad at reading. But one can certainly answer the question without that assumption. – Dylan Moreland Mar 31 '12 at 19:05
@KannappanSampath i added my answer only to show that the "geometric" fact that the reflections are precisely the elements of order 2 (when n is odd) can be shown algebraically. this is because i believe that algebraic facts should be shown algebraically when possible. – David Wheeler Mar 31 '12 at 19:37
@DavidWheeler Sure, I have no issues with that. – user21436 Mar 31 '12 at 19:42
@Dylan Added !! – user21436 Mar 31 '12 at 20:07
@Kannappan Sampath: Concerning your footnote, see the bottom of page 4 at, which includes a reference showing there are counterexamples for each odd prime $p$. – KCd Mar 31 '12 at 23:14

recall that $ba = a^{-1}b$. let's prove that $ba^k = a^{-k}b$ for all $k$, by induction on $k$:

we have the base case given above. by an induction hypothesis, we have:

$ba^{k-1} = a^{-k+1}b$, and so:

$ba^k = (ba^{k-1})a = (a^{-k+1}b)a = a^{-k+1}(ba) = a^{-k+1}(a^{-1}b) = a^{-k}b$, QED

we can use this to show that the order of $a^kb$ is $2$:

$(a^kb)^2 = (a^kb)(a^kb) = a^k(ba^k)b = a^k(a^{-k}b)b = b^2 = e$

we have exactly $n$ choices for $k$, and no element of $\langle a \rangle$ is of order $2$ (since $n$ is odd), so we have exactly $n$ elements of order $2$.

share|cite|improve this answer

Let me ignore the condition on $n$, since it isn't any more work to consider all $n$. We can think of this geometrically, or use a presentation such as \[ D_{2n} = \langle\sigma, \tau \mid \sigma^n = 1, \tau\sigma\tau = \sigma^{-1}\rangle. \] It's a good exercise to prove that every element of this group can be written uniquely as $\sigma^i\tau^j$, where $i = 0, \ldots, n - 1$ and $j = 0, 1$. When do we have $(\sigma^i\tau^j)^2 = e$? If $j = 0$ then we can write this as $\sigma^{2i} = e$, and since $\sigma$ has order $n$ either $i = 0$ (this has order $1$, so it doesn't count) or $n$ is even and $i = n/2$.

If $j = 1$, then the expression becomes $\sigma^i(\tau\sigma^i\tau) = \sigma^i\sigma^{-i} = e$ — remember that conjugation is an automorphism — regardless of $i$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.