Blogs, Symmetry

Symmetric group of an equilateral triangle.

In Symmetry of regular polygons, we found all the symmetries of regular polygons.  Today, we will look at what happens when we define the composition of such symmetries and the properties this set will have.

To begin with, recall that for the equilateral triangle we had the symmetries R0, R120, R240, F1, F2 and F3, where Ra is a rotation about the center of the triangle by an angle of a degrees, and Fa is a flip over the vertex labeled a.  As a reminder of how we labeled the vertices, we have a picture of our triangle below.  Note that the labels are number counterclockwise.

img_1849.jpg

Our goal now is to figure out what happens if we perform two of these actions in a row, so let’s look at an example.  Suppose that we rotate the triangle by 120 degrees, we then have

img_1850

We then rotate the triangle again another 120 degrees and the triangle now ends up like

img_1851

Note that if we had started out rotating 240 degrees, the resulting picture would be the same.  In this way, we say that R120oR120=R240, that is the composition of the two rotations of 120 degrees is a rotation of 240 degrees.

As another example, if we flip over the vertex labeled 1, we end up with

img_1852

If we then flip over the vertex 1 again, we end up with

img_1849.jpg

Which is the same as doing nothing, so we would say F1oF1=R0.  In order to find the composition of different symmetries, we note that while would say the left of the right these means we first put the triangle in the the right one, then put it into the second one.  To see this, we want to find R120oF1, so, we first flip the triangle over the first vertex and get

img_1852

If we now take the resulting picture and rotate it 120 degrees, we will end up with the picture

img_1853

This is the same end result as if we had flipped the original triangle over the 2 vertex, so we say that R120oF1=F2.  If we now find F1oR120, we then rotate 120 degrees and get the picture

img_1850

After this we will flip the triangle over the 1 vertex.  Note that by the 1 vertex, however, we are referring to the numbering in the original picture.  That is, the vertex 2 is now where the vertex 1 was, so this is the vertex we flip over.  Doing so, we end up with

img_1854

We no see that this is the same as flipping over the vertex 3, so we get that F1oR120=F3.  I want to point out that the reason I took the time to discuss the order of symmetries is because if we change the order in which we perform the symmetries, we end up with different answers in this case.  In the classroom with students I would do these examples, then I would leave it up to the students to find the rest of the compositions.  In order to keep track of the results, I would have the students fill out a Cayley (multiplication) table.  When using the table, the top row is the right entry and the left column is the left entry when determining the order to multiply.  Therefore, we would do the top symmetry first followed by the left symmetry.  I give the completed table below.

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Now that we have the multiplication table, we can look at some properties of this set up composition.  There a few things to note;

  1. The composition of two symmetries is again a symmetry.
  2. The composition of symmetries (AoB)oC=Ao(BoC).
  3. There is a symmetry, R0, such that R0oA=AoR0=A.
  4. For every symmetry, A, there is a symmetry, B, such that AoB=BoA=R0.
  5. There exists symmetries A and B such that AoB is not equal to BoA.

The property 1 states that the set of symmetries is closed under the binary operation of composition, 2 states that associativity holds, 3 states that an identity exists, 4 states that every symmetry has an inverse and 5 states that the composition is not commutative in general.  The first four properties tell us that these symmetries form what is called a group.  I would note that if we look at the integers under addition, then these four properties also hold where the identity is 0 and the inverse of any number k is -k since k+(-k)=0.  With the integers, however, we do have that a+b=b+a, so the fact that we don’t have commutativity in group of symmetries is something distinct that I like to point out for students.  In fact, this is the smallest example of a group that does not have the commutative property.

Together with the fact that this process allows students to actively participate in the process and play with objects, I feel this is a great way to introduce people to sets of number and operations other than addition and multiplication.  I have success with this project in general education math classes as well as upper level courses.  There is about it that provides a fun insight and different view point to students that don’t spend a lot of time with math, whereas it provides ample opportunity for math majors to pursue deeper questions.  I hope you had fun with this, and that, if you try this in class, your students enjoy it as well.

Next time in the symmetry series we will look at the symmetry groups of all polygons.  While we took the time to look at compositions of symmetries individually, but in the next post we will look at things in a more general way.  This will allow us to find what the symmetric group will look like for any polygon.  I hope you’ll join me for that journey as well in a couple days.  Make sure to follow the blog to be updated when posts are made.

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