While searching online, I've come across two ways to define the elements of the dihedral group. Both ways are internally consistent and are fine as far as I can tell, but they are mutually exclusive, so I was wondering which of the two ways is more standard or commonly used.

The two ways are as follows:

Way 1. Elements are defined as transformations against a fixed set of axes.

In this way of defining the elements, each element of $D_n$ is either a reflectional symmetry or a rotational symmetry of the polygon being considered. In an $n$-gon, there are $n$ reflectional symmetries and $n$ rotational symmetries.

The reflectional symmetries are described as follows: An $n$-gon has $n$ axes of symmetry. For each axis of symmetry in the $n$-gon there is an element in the dihedral group that reflects the $n$-gon along that axis. It is important to note that these axes stay fixed, even after the n-gon itself undergoes a rotational symmetry.

Similarly, there are $n$ rotational symmetries in the $n$-gon being discussed, and each of these rotations is a member of $D_n$. It is important to note that here, the rotations are always in the same direction, even if the shape undergoes reflectional symmetry. So, for example, an element $r$ that rotates a square $90^{\circ}$ clockwise will always rotate the square $90^{\circ}$ clockwise, even after the square is reflected.

enter image description here

Way 2. Elements are defined as permutations of vertices.

The second way of defining the elements of $D_n$ is that each element is defined as a permutation of vertices of the $n$-gon. (It should be noted that there are in total $n!$ permutations of vertices, yet only $2n$ elements in $D_n$; this discrepancy is explained by the fact that a permutation must also preserve the structure of the $n$-gon in order to be included in $D_n$.) Like all permutations on a finite set, these permutations can be written out as cycles such as $(1234)$, where $1$, $2$, $3$, and $4$ are the names of vertices of a square, for example.

enter image description here

Way 2 is actually distinct from Way 1, both geometrically and algebraically (as far as I can tell).

There are still $n$ rotations and $n$ reflections, but unlike in Way 1 where the transformations are defined in terms of a fixed "background" which doesn't move as the $n$-gon undergoes rotation or reflection, the transformations in Way 2 are now defined in terms of the vertices of the $n$-gon, which DO move as the $n$-gon undergoes rotation or reflection. What this means is that in Way 2, the transformations change depending on the current orientation of the $n$-gon. For example, consider the rotation of a square. In Way 1, we can let $r$ be the element that rotates the square $90^{\circ}$ clockwise. The direction of rotation ($90^{\circ}$ clockwise or, equivalently, $270{\circ}$ counterclockwise) never changes, even after the square is reflected across one of its axes of symmetry. On the other hand, the roughly corresponding rotation in Way 2 would be something like the cycle $(1234)$. Unlike $r$, which always rotates the square $90^{\circ}$ clockwise, $(1234)$ may rotate the square $90^{\circ}$ either clockwise or counterclockwise, depending on whether the square has been reflected or not. Similarly, the axes of reflection in Way 2 move along with the square, while those of Way 1 remain fixed.

The fact that the two ways are non-equivalent can also be verified algebraically (I think...). No isomorphism exists between the two Ways (at least as far as I can tell; I tried constructing an isomorphism by making a bijection between the elements as defined in Way 1 with the elements as defined in Way 2, matching the rotations and reflections in Way 1 with the corresponding ones in Way 2, but the bijection did not satisfy the criterion that $f(a)f(b) = f(ab)$ for all $a$, $b$ in Way 1. The problem arose when $a$ was a rotation and $b$ was a reflection. However, if there actually is an isomorphism that I overlooked, please correct me.)

My main question is: Which of these two ways is standard, or used more often by working mathematicians? Are both acceptable?

It seems to me that Way 2 is just nicer all-around, mainly because all of the elements are permutations and can thus be written as cycles, which are easy to work with algebraically (by Cayley's Theorem, Way 1 is isomorphic to some group of permutations anyway, but then it seems like kind of a hassle finding a way to write it as cycles and whatnot.)

If there are some benefits to Way 1, then I would like to learn about those too. Thanks in advance for those who bothered to read through all this!

  • 5
    $\begingroup$ Worker mathematicians represent $D_n$ as $\langle a,b \ | \ a^n = b^2 = (ab)^n = 1 \rangle$. $\endgroup$
    – Crostul
    Commented Dec 28, 2015 at 10:13
  • $\begingroup$ There is less distinction between Way 1 and Way 2 than you are claiming. In both cases, the dihedral group is being described by how it acts on some background mathematical object. In Way 1 the relevant mathematical object is the whole Euclidliean plane, and the plane does move against itself as elements of the dihedral group act on it by rotation or reflection. In Way 2 the relevant object is much smaller, namely the vertices of the polygon, but those vertrices still form a "fixed" background object which "moves" as elements of the dihedral group are applied... $\endgroup$
    – Lee Mosher
    Commented Dec 28, 2015 at 16:06
  • 1
    $\begingroup$ …The unifying idea is that a "motion" is certain kind of function from the object to itself, namely: an isometry from the Eudlicean plane to itself in Way 1; a permutation of the vertex set in Way 2. $\endgroup$
    – Lee Mosher
    Commented Dec 28, 2015 at 16:08
  • $\begingroup$ Is it actually true that no isomorphism exists between these two notations? $\endgroup$
    – Yadeses
    Commented Aug 26, 2021 at 20:45

2 Answers 2


There is a third way, which follows from the first way, but which is just as popular: $$ D_{n}=\langle x,y\mid x^{n}=y^{2}=(xy)^{2}=1\rangle $$

From this presentation, one can deduce that the elements of $D_n$ are of the form $x^i y^j$ and so $D_n$ has order $2n$ (and is commonly also denoted by $D_{2n}$, some confusion arises...)

In practice, it is quite useful to have all these various ways to think about $D_n$ (and other groups).

  • $\begingroup$ Here , usually $x^{i}y^{j}$ don't commute. So, is the action will be taken here (on $x^{i}y^{j}$) , from right to left ( I mean first reflection symmetry then rotation symmetry) or it will be taken from left to right ( mean first rotation then reflection) ? As, these two cases give different results ,, I am little bit confused here that what is actual correct order for defining this kind of element in dihedral group. $\endgroup$
    – A learner
    Commented Mar 4, 2021 at 15:00

The Way 2 by permutations is good only for small orders (say up to order $8$ or $10$). I have not seen its use for some specific purpose.

The Way 1 is geometric, and it is used many places, and also very useful in many places. For example, in regular $6$-gon (hexagon), we can (indirectly) see two regular $3$-gon (equilateral triangles). This implies that there are two copies of dihedral group of order $6$ in dihedral group of order $12$.

(The third way in an answer is better in some computation purposes.)

  • 1
    $\begingroup$ Another implication of the two triangles is that there is a homomorphism to $S_2$, the symmetric group on the triangles. Working with these ideas can bring geometric intuition to group theory. One of the copies of the dihedral group of order $6$ fixes the triangles, the other exchanges them. So this helps to understand why structures which are fixed or permuted under group actions might be worth studying. It is also not difficult to express general dihedral groups as permutation groups by reference to the positions of the vertices. $\endgroup$ Commented Dec 28, 2015 at 11:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .