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Let $G$ denote the orientation preserving isometries of Icosahedron. I want to show the following using group-theoretic notions:

Let $N\leq G$ be a subgroup with order $5.$ Show that it is a stabilizer of a vertex $v$ in the Icosahedron.

Here is my idea so far:

I know $|G|=60$. By Lagrange's theorem, the index of $N$, $$[G:N]=\frac{60}{|N|}=\frac{60}{5}=12=|V|,$$ where $V$ denotes the set of vertices of the Icosahedron. This means there is a bijection from $$G/N\to V.$$ Thus, every left coset $gN$ can be identified with a unique vertex $v\in V.$ Since the action of $G$ on $V$ is transitive (I have shown this), the claim follows.

How does this proof look? I am not quite sure whether this argument works. Please help me improve this proof!

  • Forgive my ignorance, why is $N$ normal? – angryavian Aug 11 '14 at 18:01
  • @angryavian, I don't get the question. I never said $N$ is normal. –  Aug 11 '14 at 18:03
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    Then what does $G/N$ mean? Also, even if you have a "bijection" $G/N \to V$, it does not immediately imply that the cosets of $N$ act in the way you want them to. – angryavian Aug 11 '14 at 18:05
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    I don't think this works, because you give no reason for the bijection to be related to the group action. – Jyrki Lahtonen Aug 11 '14 at 18:05
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    Consider the following fake argument. Let $G$ be the group of 24 orientation preserving symmetries of a cube. Let $N$ be a group of order two generated by a 180 degree rotation about the axis connecting the centers of two opposite vertices. $[G:N]=12$, and the cube has $12$ edges. Therefore there exists a bijection between the set of edges and the cosets of $N$ in $G$. Also $G$ acts transitively on the set of edges. By your logic this implies that $N$ is the stabilizer of an edge of the cube in $G$. Clearly this is absurd. – Jyrki Lahtonen Aug 11 '14 at 18:11
  • @JyrkiLahtonen Do you mean "the centers of two opposite faces"? – angryavian Aug 11 '14 at 18:14
  • Yes. Sorry. I failed to proofread :-) – Jyrki Lahtonen Aug 11 '14 at 18:21

2 Answers2

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Some things to consider:

  • What does the stabilizer of a vertex $v$ look like?
  • $N$ has order $5$, so what kind of group is it?
  • Consider the isometries of the icosahedron and consider which ones could possibly generate $N$.
angryavian
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  • Okay. 1) The stabilizer of a vertex $v$, ${g\in G:g(v)=v}$ is the group of rotations by angles of $\frac{2\pi}{5},\frac{4\pi}{5},\frac{6\pi}{5},\frac{8\pi}{5},2\pi$, around the axis through $v$. Thus, it a cylic group generated by rotation by $2\pi/5$ 2) If $N$ has order 5, I know it is a cylic group. 3) Rotations around a vertex generate a cyclic group of order 5. Rotations around the center of a face generate a cyclic group of order 3. Rotations around the midpoint of an edge generate a cyclic group of order 2. This means, $N$ could only be the stabilizer of a vertex. Correct? –  Aug 11 '14 at 18:32
  • @ush I think that works, good job. – angryavian Aug 11 '14 at 18:37
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    Forgive me for asking this question now: our proof only shows that $N$ could be generated by rotation around a vertex. But does it also say that it is the only way to get $N$? I want the latter. Could you help me to fill up this gap? –  Aug 11 '14 at 20:18
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    @ush $N$ is generated by an element of order $5$. You have shown that no other elements (besides the rotations about a vertex) have order $5$, so you're ok. – angryavian Aug 11 '14 at 20:48
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    I showed rotation around a vertex have order 5, and rotation around the center of an edge or a face does not have order 5. Does it immediately remove any other possibilities? For example, why can't composing two different rotations have order 5 and at the same time, this composition be different from a rotation around a vertex? –  Aug 11 '14 at 20:56
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    @ush I had assumed that you already showed that any isometry would be one of those three types of rotations, but yes you need to justify that. Then, any composition of two rotations would be another type. See page 5-6 of this pdf for more info. – angryavian Aug 11 '14 at 21:05
  • okay. can you give some hints to justify that any isometry would be one of the three types? Every element in $SO(3)$ is a rotation around a 1-dimensional linear subspace of $\mathbb{R}^3$. How to say that the subspace can only go through a vertex, or the center of an edge or a face? Geometrically, this is clear. But how to formalize this? –  Aug 15 '14 at 12:07
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One attack is to use properties of orientation preserving orthogonal linear transformation of $\Bbb{R}^3$. Those have (assuming that the center of the icosahedron is at the origin) matrices of determinant $1$ such that their transpose is also their inverse.

A group of order $5$ is necessarily cyclic. The generator $g$ of a cyclic group $\langle g\rangle=C_5\le SO(3)$ is an orthogonal transformation of the 3-space.

Prove that $g$ has an axis in $\Bbb{R}^3$. This is the eigenspace belonging to eigenvalue $1$. Observe that the axis is shared by all powers of $g$.

Observe that the point of intersection of the surface of the icosahedron and that axis is a fixed point for $\langle g\rangle$. Observe that only a vertex will work.

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Jyrki Lahtonen
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  • Is it completely obvious that the axis can only go through a vertex, or the center of an edge or that of a face? I don't see this. –  Aug 15 '14 at 12:10
  • Let $P$ be a point on the surface of the icosahedron the axis goes through. If $P$ lies on a (unique!) face, then those rotations must stabilize the said face. As the face is a triangle this disallows a cyclic group of order five. Similarly if $P$ lands on a (unique!) edge. Ergo, it has to be vertex. – Jyrki Lahtonen Aug 15 '14 at 12:17
  • But I don't know whether $N$ is a stabilizer in the first place. I only know $N\leq G$ and $|N|=5.$ –  Aug 15 '14 at 12:22
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    The point of my answer was to show that if you have a subgroup $N$ isomorphic to $C_5$ inside $SO(3)$, then that subgroup consists of five rotations - all about the same axis. Thus $P$ is a fixed point for $N$. – Jyrki Lahtonen Aug 15 '14 at 12:30