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

I know we can generate dihedral group of order three ($D_3$) and four ($D_4$) but my question is whether we can generate dihedral group of order five?

share|cite|improve this question
If you mean "can we define a dihedral group of order five", then you should realise that the dihedral group of order three is the symmetries of a triangle while the dihedral group of order four is the symmetries of a square. A pentagon has got some symmetries, so, well, we might as well call this the dihedral group of order five. I mean, why not?! – user1729 Mar 12 '13 at 10:40
You should try and use standard terminology. The order of a group is the number of its elements, so in fact there is no dihedral group of order 3 and none of order 5. And nobody knows what you mean by "generate a group", so please explain! – Derek Holt Mar 12 '13 at 11:11
up vote 5 down vote accepted

Writing $D_n$ for the group of symmetries of a regular $n$-sided polygon, we have $$D_n=\langle\,a,b\mid a^n=1,b^2=1,ba=a^{-1}b\,\rangle$$ that is, $D_n$ is generated by the two elements $a$ and $b$ subject to the relations $a^n=1$, $b^2=1$, and $ba=a^{-1}b$.

We can identify $a$ with a rotation through $2\pi/n$, and $b$ with a flip in any axis of symmetry.

If this doesn't answer the question ... then please edit the question to give us a better idea of what's wanted.

share|cite|improve this answer

There are dihedral groups for every regular polygon with $n$ vertices. Each one has order $2n$, respectively, so all of them have even group order. If you are using "order" to refer to the number of vertices of the polygon, then "yes".

share|cite|improve this answer

$D_3$ has order six, not three, and $D_4$ has order eight, not four. In general, $D_{n}$ has order $2n$ - that's one cyclic group of rotations (there are $n$ vertices, and you rotate by one vertex each time) and the same number of reflections (across each of the $n$ diagonals and faces). Perhaps it would help for you to draw these out with a pentagon and hexagon.

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.