This is something I don't understand. I've read that a permutation is even if it can be arrived by an even number of transpositions. OK, but then let's consider the dihedral group D3 for an equilateral triangle.

For the triangle, the permutations are: the identity transformation, two 120 degree rotations, and 3 flips about the axis of symmetry. What I don't get is why the two 120 degree rotations are considered of even parity. Could someone please explain to me? The identity transformation is easy enough because it requires 0 transformations to get there.

Are the 3 flips about the axis of symmetry even as well? After all, they can be represented in the Cayley table, which consists of 2 transformations as well.

  • 1
    $\begingroup$ In addition to what folks have said: you seem to be confusing transformation (which is a more abstract notion here) with transposition, which is specifically the interchange of two elements (and is specific to the symmetry group/permutation group). The parity of a permutation is the number of interchanges of elements - the number of transpositions - necessary to get the permutation; it's a theorem, more than a definition, that this concept is even well-defined! $\endgroup$ – Steven Stadnicki May 7 '18 at 17:14

The $D_3$ group is seen as a set of transformations that act on the vertices of the triangle.

If we label the vertices $1$,$2$ and $3$, then the $120$ degree rotations can be written in permutation form as $(1\;2\;3)$ and $(1\;3\;2)$ respectively. But we have that

$$(1\;2\;3)=(1\;3)(3\;2) \qquad \text{and} \qquad (1\;3\;2)=(1\;2)(2\;3)$$

Hence these are even permutations.

  • $\begingroup$ Am I reading that cycle notation wrong or do you have those reversed? In other words, doesn't $(1\, 2\, 3) = (1\, 2)(2\, 3)$ and $(1\, 3\, 2) = (1\, 3)(3\, 2)$? Ohh... I was taught a convention where you read cycle notation right to left, but you are reading/writing it left to right? $\endgroup$ – Todd Wilcox May 7 '18 at 15:51
  • $\begingroup$ Left to right is very common. It makes no odds. $\endgroup$ – ancientmathematician May 7 '18 at 15:53
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    $\begingroup$ @ancientmathematician I'm not familiar with the phrase, "it makes no odd". I wonder why I was taught right to left. Anyway, I guess if I put cycle notation anywhere, and intend it to be read right to left, I better specify. $\endgroup$ – Todd Wilcox May 7 '18 at 15:55
  • $\begingroup$ Some people prefer right to left. It really doesn't matter provided one is consistent. $\endgroup$ – ancientmathematician May 7 '18 at 15:57
  • $\begingroup$ Related: math.stackexchange.com/questions/527514/… $\endgroup$ – Todd Wilcox May 7 '18 at 16:04

Suppose that the triangle has one red side (initially up, say) and one blue side (initially down). When you flip about an axis of symmetry, that will always change the colour of which side is up. Those are odd permutations.

Now suppose you do two flips in a row. The effect on the colour is that you'll be back to where you were at the beginning. These are even permutations. But will it be the identity transformation?

Not necessarily: only if the two flips were in the same axis of symmetry. The other times you will have just executed one of the two $120$ degree rotations!

A permutation is just a bunch of independent cycles. A cycle is an odd permutation if it has an even number of elements, and it's even if it has an odd number of elements. This is exactly what you find with $D_3$ here. And a permutation is even exactly if it has an even number of odd cycles.

Just play around with some small groups like you are doing until you feel comfortable with this fundamental idea. Hope this helps! Have fun!

  • $\begingroup$ But what about the flips from the 3 lines of symmetry? Are they even or odd permutations? If they can be inside the Cayley table, then it must reason that 2 transformations led them there and hence they are even? $\endgroup$ – Yip Jung Hon May 7 '18 at 14:07
  • $\begingroup$ These flips are odd. They are in the Cayley table, and are the result of odd perm followed by an even perm which gives an odd perm. (Or even followed by odd, which again gives odd.) $\endgroup$ – Laska May 7 '18 at 14:21

A 120 degree rotation sends $(A,B,C)$ to $(B,C,A)$ and thus is the same as $(A,B,C) \mapsto (B,A,C)$ followed by $(B,A,C) \mapsto (B,C,A)$, two transpositions.


There are several different characterizations of permutations that are equivalent:

  1. Are composed of an even number of transpositions
  2. Are composed of cycles of odd number of elements
  3. Are member of the alternating group $A_n$
  4. Matrix representation has positive determinant
  5. Preserve orientation

If you're viewing the permutation group as acting geometrically, the last one is more salient than otherwise. If you look at where all the points are relative to each other, the identity and the two 120° rotations leave these relationships unchanged. Suppose you form a triangle on a clock face, with a red dot at 12, green at 4, and blue at 8. The three even permutations result in a configuration such that going around clockwise from red, you get green and then blue. The odd permutations result in configurations in which, starting at red and going clockwise, you have blue and then green.

Equivalently, the identity and the two 120° rotations can all be performed within the space that the triangle is embedded in; a triangle can be rotated while remaining in the plane. To physically reflect a triangle, however, requires rotation it through three dimensional space, rather than just the plane. In general, permutations of $n$ elements can be represented as actions on vertices of an $n-1$ regular simplex in $n-1$ space, where even permutations correspond to rigid transformations within that $n-1$ space, while odd permutations correspond to permutations that require $n$ dimensions to perform.


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