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Group of symmetries on a regular polygon 
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#1
Oct2607, 12:29 PM

P: 176

So i began reading up on some group theory and I came across an interesting question, what is the order of the group of symmetries on of a nsided regular polygon?
with a square it's 8, triangle it's 4. I feel like i'm missing something with the pentagon cuz i'm only finding these: the 5 rotations, two diagonal reflections which are NOT the same as that for the square, reflection over the vertical axis. i'd appreciate any casual discussion on the topic as I find it fascinating, at the very beginning because symmetries behave very much like permutations (if we label vertices) I thought there might be a relationship between this set and the symmetric group of {1,...n}? But there are obviously permutations that no movement of a polygon in the plane can mimic. true or false: the group of symmetries of a nsided regular polygon in the plane is isomorphic to a subset of the symmetric group of {1,...,n}. cheerio! 


#2
Oct2607, 01:43 PM

P: 1,076

There are more than 4 for a triangle. I'm also not sure how you can compare the diagonal reflections of a pentagon with those of a square, but there are more than 2.
Look at the triangle again, and try to find all of it's symmetries (label the vertices with 1,2,3, and see f that helps you find more) then try to make an inference as to the general order of this group based on your findings for the triangle and the square. 


#3
Oct2607, 01:59 PM

P: 176

there are two more for the triangle  with the base of the triangle on a horizontal axis we can switch the top vertices with the right vertices by flipping the triangle from the left vertices and vice versa with if you flip the triangle from the right vertices.
2n? 


#4
Oct2607, 07:50 PM

P: 40

Group of symmetries on a regular polygon
http://en.wikipedia.org/wiki/Dihedral_group should be of use to you. In direct response to your question, the page notes that for small n, the Dihedral group isn't a subgroup of the symmetric group S_n, as demonstrated by the fact n! < 2n for n=1.



#5
Oct2607, 09:29 PM

P: 291

Here is how to think of it.
1)Draw a regular polygon with vertices 1,2,...,n in order. 2)The question is how many ways can the polygon be replaced by rigid motions, i.e. with vertices still in order. Now for any one of the n locatations there are n choices. Having chosen that there is only one chose for the next one in succession, i.e. either left or right. Having chosen that the polygon is completely determined. Thus there are 2n elements in this group. 3)Let a be the positive rotation by 2pi/n i.e. the cycle (1,2,...,n) and let b be the reflection through the middle and vertex 1, i.e. (2,n)(3,n1)..... 4)So a^n = 1 and b^2 = 1. 5)Consider S = {a^k,ab^k} for k=1,2,...,n 6)There are 2n elements in that set all of which are distinct. So S represents the group D_n. 7)Now it remains to show ba=a^{n1}b which is not so hard to show. Hence the group presentation for the dihedral group is: [tex]\left< a,b a^n = 1, \ b^2 = 1, \ ba=a^{n1}b \right>[/tex] 


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