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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 reflections2012-01-27
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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.2012-01-27
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    @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 it2012-01-27
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    Anton, are you preparing to exam on the geometry course by Sossinsky in Independent University of Moscow?2012-01-27
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    Hi, Sergey, actually I am not :)2012-01-30

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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.2012-01-27
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    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 $h$. @JackSchmidt2012-01-27
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    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?2012-01-27
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    @ThomasAndrews: perfect, thanks. "being opposite is a geometric property"2012-01-27
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    (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$2012-01-27
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    @Anton: yup, except you probably want two conjugacy classes each of size 1. S5 only has one of those, the identity. Who is the other mysterious element?2012-01-27
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    @Anton More elementary, I'd just note that the only elements of order $2$ in $S_5$ must be either transpositions of a pair $(a b)$ or a product of disjoint transpositions, $(a b)(c d)$. Then show that the groups generated by such elements can't be normal. (Hint: There is always at least one element fixed by an element of order 2 in $S_5$.)2012-01-27
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    In particular, if $\sigma$ is an element of order two in $S_n$ for $n\geq 3$, then then since $\sigma\neq 1$, there is an $a\in\{1,...,n\}$ such that $\sigma(a)\neq a$ and a $b\in\{1,2,...n\}$ so that $a\neq b$ and $\sigma(a)\neq b$. Let $\alpha\in S_n$ be the transposition $(a b)$. Then $$\alpha\sigma\alpha^{-1}(b) = \alpha\sigma(a) = \sigma(a) \neq a = \sigma(b)$$ So $\alpha\sigma\alpha^{-1}\neq \sigma$ so $\{1,\sigma\}$ cannot be a normal subgroup of $S_n$.2012-01-27
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    Whoops, typo in that last argument. That last line should be:$$\alpha\sigma\alpha^{-1}(b) = \alpha\sigma(a) = \sigma(a) \neq \sigma(b)$$2012-01-27
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    @ThomasAndrews: Thanks a lot!2012-01-27
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    @JackSchmidt: the mysterious element? could you be thinking of -1?2012-01-27
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    @Anton: in the (full) icosahedral group yes, but in S5 there is no such element, just like Thomas Andrews said.2012-01-27
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    @JackSchmidt: Thanks! I have a very stuid question: What exactly is a symmetry? In my understanding it is something we can do to a figure/object so that it is mapped to its original form without breaking the edges joining the vertices. Would this be right? If so, what is the -1 symmetry doing? if it is the g Thomas mentioned ie it sends all the points to their oposite points does the figure get all tangled up? and what plane is the reflection done wrt?2012-01-27
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    @Anton: Your understanding is good (that is called a combinatorial, or graph symmetry). Thomas Andrews in his second reply to me used the understanding "a mapping that never changes distance" (a metric isometry). The -1 symmetry is a very special feature of 3D; a "super-reflection" formed from 3 reflections in the XY-, XZ-, and YZ- planes. You can think of it as turning something inside out. It is also called an "inversion". It is not the reflection over any plane, but it is the only non-rotation that is not a reflection+rotation.2012-01-27
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    @JackSchmidt: Thanks! :)2012-01-27
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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?2012-01-27
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    Perhaps we can show that the group of all its symmetries must be a direct product?2012-01-27