group actions群作用ぐんさよう and the orbit-stabilizer theorem軌道固定部分群定理きどうこていぶぶんぐんていり: basic exercises

1Corresponding lecture

2Problem 1: the left regular action

For a group $$G$$G, define $$g·x=gx$$g\cdot x=gx. Prove that this is an action of $$G$$G on itself, and find the orbit and stabilizer of $$x\in G$$x\in G.

2.1Answer

We have $$e·x=ex=x$$e\cdot x=ex=x and $$\left(gh\right)·x=\left(gh\right)x=g\left(hx\right)=g·(h·x)$$(gh)\cdot x=(gh)x=g(hx)=g\cdot(h\cdot x), so the action axioms hold. Every $$y\in G$$y\in G can be written as $$y=\left(y{x}^{-1}\right)·x$$y=(yx^{-1})\cdot x, so the orbit is all of $$G$$G. If $$g·x=x$$g\cdot x=x, right cancellation gives $$g=e$$g=e, so the stabilizer is $$\left\{e\right\}$$\{e\}.

3Problem 2: vertices of an equilateral triangle

The six symmetries of an equilateral triangle act on its three vertices. Find the sizes of the orbit and stabilizer of a vertex $$x$$x, and verify the orbit-stabilizer theorem.

3.1Answer

Every vertex is reachable, so $$\left|Gx\right|=3$$|Gx|=3. The identity and the reflection across the axis through $$x$$x fix the vertex, so $$\left|G_x\right|=2$$|G_x|=2. Therefore

4Problem 3: kernel of an action and stabilizers

For the homomorphism $$\rho :G\to \mathrm{Sym}\left(X\right)$$\rho:G\to\operatorname{Sym}(X) associated with an action, prove

Then explain why the action on the vertices of an equilateral triangle has $$\ker \rho =\left\{e\right\}$$\ker\rho=\{e\} although $$\left|G_x\right|=2$$|G_x|=2.

4.1Answer

An element $$g$$g lies in $$\ker \rho$$\ker\rho exactly when $$\rho \left(g\right)$$\rho(g) is the identity map, equivalently when $$g·x=x$$g\cdot x=x for every $$x\in X$$x\in X. This is exactly the condition that $$g$$g lie in every $$G_x$$G_x.

For the triangle, only the identity fixes all three vertices, so the kernel is trivial. A reflection through the axis containing one chosen vertex also fixes that vertex, so its stabilizer has two elements.