Subgroups of order $5$ in Icosahedral group

Question

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!

Answer 1

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 $N$.

Answer 2

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.