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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?

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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

3 Answers 3

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.

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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".

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$D_3$ 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.

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