If a Finite Group Acts on a Set Freely and Transitively, then the Numbers of Elements are the Same

Group Theory Problems and Solutions in Mathematics

Problem 488

Let $G$ be a finite group and let $S$ be a non-empty set.
Suppose that $G$ acts on $S$ freely and transitively.
Prove that $|G|=|S|$. That is, the number of elements in $G$ and $S$ are the same.

 
LoadingAdd to solve later

Definition (Free and Transitive Group Action)

  • A group action of a group $G$ on a set $S$ is called free if whenever we have
    \[gs=hs\] for some $g, h\in G$ and $s\in S$, this implies $g=h$.
  • A group action of a group $G$ on a set $S$ is called transitive if for each pair $s, t\in S$ there exists an element $g\in G$ such that
    \[gs=t.\]

Proof.

We simply denote by $gs$ the action of $g\in G$ on $s\in S$.

Since $S$ is non-empty, we fix an element $s_0 \in S$. Define a map
\[\phi: G \to S\] by sending $g\in G$ to $gs_0 \in S$.
We prove that the map $\phi$ is bijective.

Suppose that we have $\phi(g)=\phi(h)$ for some $g, h\in G$.
Then it gives $gs_0=hs_0$, and since the action is free this implies that $g=h$.
Thus $\phi$ is injective.

To show that $\phi$ is surjective, let $s$ be an arbitrary element in $S$.
Since the action is transitive, there exists $g\in G$ such that $gs_0=s$.
Hence we have $\phi(g)=s$, and $\phi$ is surjective.

Therefore the map $\phi:G\to S$ is bijective, and we conclude that $|G|=|S|$.


LoadingAdd to solve later

More from my site

  • Finite Group and a Unique Solution of an EquationFinite Group and a Unique Solution of an Equation Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that \[b^m=a.\]   We give two proofs. Proof 1. Since $m$ and $n$ are relatively prime […]
  • The Order of a Conjugacy Class Divides the Order of the GroupThe Order of a Conjugacy Class Divides the Order of the Group Let $G$ be a finite group. The centralizer of an element $a$ of $G$ is defined to be \[C_G(a)=\{g\in G \mid ga=ag\}.\] A conjugacy class is a set of the form \[\Cl(a)=\{bab^{-1} \mid b\in G\}\] for some $a\in G$. (a) Prove that the centralizer of an element of $a$ […]
  • $p$-Group Acting on a Finite Set and the Number of Fixed Points$p$-Group Acting on a Finite Set and the Number of Fixed Points Let $P$ be a $p$-group acting on a finite set $X$. Let \[ X^P=\{ x \in X \mid g\cdot x=x \text{ for all } g\in P \}. \] The prove that \[|X^P|\equiv |X| \pmod{p}.\]   Proof. Let $\calO(x)$ denote the orbit of $x\in X$ under the action of the group $P$. Let […]
  • Group of Invertible Matrices Over a Finite Field and its StabilizerGroup of Invertible Matrices Over a Finite Field and its Stabilizer Let $\F_p$ be the finite field of $p$ elements, where $p$ is a prime number. Let $G_n=\GL_n(\F_p)$ be the group of $n\times n$ invertible matrices with entries in the field $\F_p$. As usual in linear algebra, we may regard the elements of $G_n$ as linear transformations on $\F_p^n$, […]
  • Inverse Map of a Bijective Homomorphism is a Group HomomorphismInverse Map of a Bijective Homomorphism is a Group Homomorphism Let $G$ and $H$ be groups and let $\phi: G \to H$ be a group homomorphism. Suppose that $f:G\to H$ is bijective. Then there exists a map $\psi:H\to G$ such that \[\psi \circ \phi=\id_G \text{ and } \phi \circ \psi=\id_H.\] Then prove that $\psi:H \to G$ is also a group […]
  • A Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is NormalA Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is Normal Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer. Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$. Then prove that $H$ is a normal subgroup of $G$. (Michigan State University, Abstract Algebra Qualifying […]
  • Nontrivial Action of a Simple Group on a Finite SetNontrivial Action of a Simple Group on a Finite Set Let $G$ be a simple group and let $X$ be a finite set. Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$. Then show that $G$ is a finite group and the order of $G$ divides $|X|!$. Proof. Since $G$ acts on $X$, it […]
  • Subgroup of Finite Index Contains a Normal Subgroup of Finite IndexSubgroup of Finite Index Contains a Normal Subgroup of Finite Index Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.   Proof. The group $G$ acts on the set of left cosets $G/H$ by left multiplication. Hence […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Group Theory
Group Theory Problems and Solutions in Mathematics
Every Group of Order 72 is Not a Simple Group

Prove that every finite group of order $72$ is not a simple group.

Close