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Proving that the Dihedral Groups are non-cylic

  1. Oct 18, 2014 #1
    Hello everyone,

    I am suppose to show that all the Dihedral groups (##D_n##, for ##n >2##) are noncyclic. I know that every cyclic group must be abelian. So, what I intended on showing was that at least two elements in ##D_n## are not commutative. Here are my thoughts:

    Because we are dealing with rigid transformations of an n-gon, any transformations must preserve the distance between any two points on the n-gon. We are particularly interested in the preservation of the distances between the vertices. In our book we are told that every transformation on a n-gon takes that n-gon to itself as a rigid object either by a rotation or reflection.

    Having already worked with ##D_3## and ##D_4##, I know that, in general, the transformation of flip and rotation are not commutative. So, I was going to try and determine an arbitrary rotation and flip and show that they are not commutative.

    If we label the vertices with the numbers ##1,2,3...,n##, then we can represent the initial configuration of the n-gon as a n-tuple: ##(1,2,3,...,n)##.

    Whenever I perform a rotation of any sort, every one of the vertices must move. To put it another way, if ##i## and ##j## are neighboring vertices, and a rotation causes ##i## to move to the vertex position ##i+1##, then ##j \rightarrow j+1##.

    Here is something else I noticed when working with ##D_4##. There is a "basic angle" through which the polygon can rotate. In the case of ##D_4##, the "basic angle" was ##R_{90}##; and all of the other rotations were built off of this “basic angle, for instance, ##R_{180}## and ##R_{270}##. This “basic angle” always moved the position of an element in the tuple by one position.

    This brings up another observation: the only permitted rotations are those which are some multiple of the interior angle (in the case of ##D_4##, this would be ##\frac{\pi}{2}##). If this generalizes to any n-gon, then the “basic angle” through which you could rotate it would be ##\frac{(n-2) \pi}{2n}##. We could denote this as ##R_{\frac{(n-2) \pi}{2n}}##.

    Now, when I rotate the n-gon by an angle of ##\frac{(n-2) \pi}{2n}##, every vertex would move over one position; that is,

    ##R_{\frac{(n-2) \pi}{2n}} ~:~ (1,2,3,...,n) \rightarrow (n,1,2,3,…,n-1)##,

    where ##(1,2,3,…,n)## is the initial configuration of the n-gon. Now, here is the trouble emerges: what would a flip look like, in terms of n-tuples?

    Sorry that these thoughts are scattered all about. Hopefully my thoughts make some sense.
     
  2. jcsd
  3. Oct 18, 2014 #2

    Zondrina

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    The dihedral group of degree ##n > 2## and order ##2n##, which I denote ##D_n##, can be expressed:

    $$D_n = <x, a \space | \space a^n = x^2 = e, xax^{-1} = a^{-1}>$$

    For ##n > 2##, the elements ##a## and ##x## do not commute, where ##a## is a rotation and ##x## a reflection.

    If you try to define a flip using an n-tuple like you have, then you're going to need to treat even and odd groups differently to define a flip. For odd groups you can define a flip:

    ##(1, 2, 3, 4, ... , n) \rightarrow (1, n, n-1, ..., 4, 3, 2, 1)##

    For example, taking ##D_5##:

    ##(1, 2, 3, 4, 5) \rightarrow (1, 5, 4, 3, 2)##

    Can you find the even case?
     
    Last edited: Oct 18, 2014
  4. Oct 20, 2014 #3
    Is the reason why we have to divide the problem into cases, even and odd, have to do with the axis of symmetry (AOS), the line about which you rotate the n-gon? In the case of an even amount of vertices, the AOS intersects two vertices, and when you rotate the n-gon about this line, these two vertices will retain their position in the n-tuple; in the case of the an odd amount of the vertices, the AOS doesn't always have to intersect two vertices. Is this right?

    So, I took the example of a square, labelling the initial configuration as ##(1,2,3,4)##, and considered a flip about the line through the vertices 1 and 3. Labelling this transformation as ##F_\nearrow##, the transformation changes the 4-tuple as such:

    ##F_\nearrow : (1,2,3,4) \rightarrow (1,4,3,2)##.

    Roughly, if the ##i##-th position is an odd number, then ##F_\nearrow : i \rightarrow i##; if the ##i##-th position is even, then ##F_\nearow : i \rightarrow i+2##. In general, then, a flip about the line connecting 1 and 3 would look like

    ##(1,2,3,4,...,n-2,n-1,n) \rightarrow (1,4,3,2,...,n,n-1,n-2)##

    Does this seem right?
     
    Last edited: Oct 20, 2014
  5. Oct 20, 2014 #4
    Hmm, I know what I did in the previous post was wrong...

    I tried finding all of the flip transformations ##D_6##, and I noticed that I can't even flip it about the line connecting 1 and 3, as this would not constitute a rigid as defined in my first post.
     
  6. Oct 20, 2014 #5

    Zondrina

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    It appears defining a flip is a bit more complicated that I had thought of originally. Looking at a few symmetries:

    18140.jpg

    Suppose you label the nodes counter-clockwise. So for the pentagon I would start at the top.

    Draw a line that cuts each shape along its axis of symmetry. The line will touch one node or two nodes depending on the shape. For a triangle it would be one node. A square would be two. A pentagon would be one. A hexagon would be two and so on. So for the odd groups it appears that one node is always on the line. For even groups there are two nodes.

    We observe the flip from earlier for the odd case since it easy to see the symmetry. The even case is different. Take the same line of symmetry for each even group. So if you connect the bottom left corner of the square to the top right corner, you would connect the bottom left hexagon node to the top right node.

    Is there any noticeable pattern?
     
  7. Oct 21, 2014 #6

    pasmith

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    That's more than is necessary: you only have to exhibit a specific rotation and a specific reflection and show that they do not commute.

    Label the vertices of the [itex]n[/itex]-gon anti-clockwise from 0 to [itex]n-1[/itex]. Then there is a rotation [itex]\rho: k \mapsto k + 1[/itex] using addition modulo [itex]n[/itex], corresponding to rotation anti-clockwise through an angle of [itex]2\pi/n[/itex]. This rotation generates all other rotations, which are of the form [itex]\rho^a : k \mapsto k + a[/itex] for constant [itex]a \in \{0, \dots, n-1\}[/itex] and correspond to rotation anti-clockwise through [itex]2\pi a/n[/itex].

    Whether [itex]n[/itex] is even or odd there is also a reflection [itex]m : k \mapsto n - k[/itex], again using addition modulo [itex]n[/itex], corresponding to the axis of symmetry through the vertex 0.

    It is straightforward to obtain expressions for [itex]m(\rho(k))[/itex] and [itex]\rho(m(k))[/itex] and show that these are not congruent mod [itex]n[/itex] for [itex]n > 2[/itex].


    EDIT: The other way to show that [itex]D_n[/itex] is not cyclic for [itex]n > 2[/itex] is to show that [itex]D_n[/itex] does not contain any elements of order [itex]2n[/itex]. Geometrically such an element would have to be a rotation through [itex]\pi/n[/itex]. Is that a symmetry of a regular [itex]n[/itex]-gon?
     
    Last edited: Oct 21, 2014
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