753: For Group Action, Induced Map with Fixed Group Element Is Bijection

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for group action, induced map with fixed group element is bijection

---

Topics

About: group
About: set

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

---

Starting Context

• The reader knows a definition of group action.
• The reader knows a definition of bijection.

---

Target Context

• The reader will have a description and a proof of the proposition that for any group action, the induced map with any fixed group element is a bijection.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$G$: $\in \{\text{ the groups }\}$
$S$: $\in \{\text{ the sets }\}$
$f$: $:G\times S\to S$, $\in \{\text{ the group actions }\}$
$g\in G$
$f_g$: $:S\to S,s\mapsto f(g,s)$
//

Statements:
$f_g\in \{\text{ the bijections }\}$
//

---

2: Natural Language Description

For any group, $G$, any set, $S$, any group action, $f:G\times S\to S$, and any element, $g\in G$, $f_g:S\to S,s\mapsto f(g,s)$ is a bijection.

---

3: Proof

Whole Strategy: Step 1: choose each $s_1,s_2\in S$ such that $s_1\ne s_2$, suppose that $f_g(s_1)=f_g(s_2)$, and find a contradiction; Step 2: choose each $s\in S$ and see that $f_g(f(g^{-1},s))=s$.

Step 1:

Let us see that $f_g$ is an injection.

Let $s_1,s_2\in S$ be any such that $s_1\ne s_2$.

Let us suppose that $f_g(s_1)=f_g(s_2)$.

$g{s}_1=g{s}_2$. $s_1=1s_1=(g^{-1}g)s_1=g^{-1}(g{s}_1)=g^{-1}(g{s}_2)=(g^{-1}g)s_2=1s_2=s_2$, a contradiction.

So, $f_g(s_1)\ne f_g(s_2)$, and $f_g$ is an injection.

Step 2:

Let $s\in S$ be any.

$f_g(g^{-1}s)=g(g^{-1}s)=(g{g}^{-1})s=1s=s$.

So, $f_g$ is a surjection.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>