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- Apr 14, 2013

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Hey!!

Let $G$ be a permutation group of a set $X\neq \emptyset$ and let $x,y\in X$. We define:

\begin{align*}&G_x:=\{g\in G\mid g(x)=x\} \\ &G_{x\rightarrow y}:=\{g\in G\mid g(x)=y\} \\ &B:=\{y\in X\mid \exists g\in G: g(x)=y\}\end{align*}

Show the following:

I have done the following:

We have to show the properties of a subgroup.

We have to show that every element of $G$ is in exactly one of these subsets, right? How can we do that?

Since $g\in G_{x\rightarrow y}$ we have that $g\in G\mid g(x)=y$.

Since $u\in G_x$ we have that $u\in G\mid u(x)=x$.

Since $v\in G_{x\rightarrow y}$ we have that $v\in G\mid v(x)=y$.

Isn't obvious that $g\circ u\in G_{x\rightarrow y}$ and $g^{-1}\circ v\in G_x$, since we look always at the outer left function?

The is a hint given that the proof is similar to the idea to show that theere are $48$ symmetries of a cube.

How exactly can we use this?

Let $G$ be a permutation group of a set $X\neq \emptyset$ and let $x,y\in X$. We define:

\begin{align*}&G_x:=\{g\in G\mid g(x)=x\} \\ &G_{x\rightarrow y}:=\{g\in G\mid g(x)=y\} \\ &B:=\{y\in X\mid \exists g\in G: g(x)=y\}\end{align*}

Show the following:

- $G_x$ is a subgroup of $G$.
- The set $\{G_{x\rightarrow y}\mid y\in B\}\subseteq 2^G$ is a partition of $G$.
- Let $g\in G_{x\rightarrow y}$ and let $u\in G_x$ and $v\in G_{x\rightarrow y}$. Then it holds that $g\circ u\in G_{x\rightarrow y}$ and $g^{-1}\circ v\in G_x$.
- The maps \begin{align*}\alpha:G_x\rightarrow G_{x\rightarrow y}, \ u\mapsto g\circ u \\ \beta: G_{x\rightarrow y}\rightarrow G_x, \ v\mapsto g^{-1}\circ v\end{align*} are to each other inverse bijections.
- Let $G$ be finite. Then it holds that $|G|=|B|\cdot |G_x|$.

I have done the following:

**For 1:**We have to show the properties of a subgroup.

**For 2:**We have to show that every element of $G$ is in exactly one of these subsets, right? How can we do that?

**For 3:**Since $g\in G_{x\rightarrow y}$ we have that $g\in G\mid g(x)=y$.

Since $u\in G_x$ we have that $u\in G\mid u(x)=x$.

Since $v\in G_{x\rightarrow y}$ we have that $v\in G\mid v(x)=y$.

Isn't obvious that $g\circ u\in G_{x\rightarrow y}$ and $g^{-1}\circ v\in G_x$, since we look always at the outer left function?

**For 5:**The is a hint given that the proof is similar to the idea to show that theere are $48$ symmetries of a cube.

How exactly can we use this?

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