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Could someone please explain the difference between the group of all icosahedral symmetries and S5? I know that the former is a direct product, but don't they work the same? Say I have an icosahedron, why wouldn't S5 work as a description of its symmetries? Thank you very much.

Added: When counting the symmetries of a platonic solid, in this case the icosahedron, Does it include reflecting along a plane cutting through the solid, in a sort of turing itself inside out reflection? I read that the symmetries counted should be "orientation-preserving". What does that mean?

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If you don't even know how to define the group of icosahedral symmetries, how are you going to answer this question? Presumably, if this is a question from somewhere, then it probably included reflections, since there are only 60 icosahedral symmetries without reflections, and 120 with reflections, and $|S_5|=120$, so the question would be trivial without reflections –  Thomas Andrews Jan 27 '12 at 16:06
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(Orientation-preserving is the same as not counting reflections, yes. There are a number of different definitions for preserving orientation, but one meaning is by representing your symmetry as a matrix. Then it is "orientation-preserving" if the determinant of the matrix is positive. –  Thomas Andrews Jan 27 '12 at 16:14
    
@ThomasAndrews : You are quite right. I was thinking that the question is a little strange since there are only 60 rotational symmetries. As to why S5 is not a right representation of the group of icosahedral symmetries, I was told that there is a simple reason as to why it can't be true without knowing that the group is isomorphic to $A_5\times \{\pm 1\}$, but I still don't see it –  Anton Jan 27 '12 at 16:14
    
Anton, are you preparing to exam on the geometry course by Sossinsky in Independent University of Moscow? –  Sergey Filkin Jan 27 '12 at 19:28
    
Hi, Sergey, actually I am not :) –  Anton Jan 30 '12 at 21:16

2 Answers 2

up vote 4 down vote accepted

Hint: If $g$ is the symmetry that send each point to it's opposite point in the icosahedron, then show $\{1,g\}$ is a normal subgroup of the group of symmetries. Show that $S_5$ does not have any normal subgroup of order $2$.

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This is very easy to do with matrices (the matrix is -1). Is it easy to see from a more elementary idea of symmetry? I guess one only needs to see it commutes with rotations. I teach a freshmen non-stem course, and they would not understand the matrix argument. –  Jack Schmidt Jan 27 '12 at 16:28
    
If $h$ is a symmetry, then $h$ sends opposite points to opposite points. This means that $g(h(x))=h(g(x))$ - the point oppsite $h(x)$ is the image under $h$ of the point opposite to $x$. But this means that $gh=hg$, for any $h$. @JackSchmidt –  Thomas Andrews Jan 27 '12 at 16:34
    
Thanks, Thomas. Can I prove the second bit by saying that a normal subgroup must be a union of conjugacy classes and there is no combination of ccl in S5 that has only 2 elements? –  Anton Jan 27 '12 at 16:40
    
@ThomasAndrews: perfect, thanks. "being opposite is a geometric property" –  Jack Schmidt Jan 27 '12 at 16:41
    
(This fact that any $h$ must commute with $g$ is "intuitive," but we can also see it by the fact that $h$ is an isometry, and, for any vertex $x$ of the icosahedron, the there is exactly one point $y$ on the icosahedron with $d(x,y)=D$, where $D$ is the diameter of the icosahedron, and that $y$ is $g(x)$. So if $x$ is a vertex and $h$ any symmetry, then $d(h(x),h(g(x)))=d(x,g(x))=D$. So $h(g(x))$ is of distance $D$ from $h(x)$, which is uniquely $g(h(x))$, so $h(g(x))=g(h(x))$ for all vertices $x$, and hence $hg=hg$ –  Thomas Andrews Jan 27 '12 at 16:42

Two groups are treated as same if there is an isomorphism between them.A simple reason why $S_5$ cannot be used to describe the symmetries of an icosahedron(whose group of symmetries we will call $I_h$) is that its structure is fundamentally different from that of $I_h$. For starters, $S_5$ cannot be expressed as a direct product of two groups unlike $I_h$, for which it is possible to do so. So $S_5$ cannot be isomorphic to $I_h$.

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Thanks, Fortuon, is there an argument that does not involve knowing the group $I_h$ beforehand? –  Anton Jan 27 '12 at 16:16
    
Perhaps we can show that the group of all its symmetries must be a direct product? –  Anton Jan 27 '12 at 16:19

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