Team 3 (George, Mackenzie, Bradley, Joe, Scott, Jordan)

Group Theory

Group Action!

A group action is a way of describing symmetries of objects using groups. The essential elements of the object are described by a set and the symmetries of the object are described by the symmetry group of this set, which consists of bijective transformations of the set."Group actions generalize group multiplication. If G is a group and X is an arbitrary set, a group action of an element $g\in$G and $x\in$X is a product,$gx$, living in X. Many problems in algebra may best be attacked via group actions. For example, the proofs of the Sylow theorems and of Burnside's Counting Theorem are most easily understood when they are formulated in
terms of group actions."[1]

Definition: Group Acting on Sets

Let $X$ be a set and $G$ be a group. A (left) action of $G$ on $X$ is a map $G \times X \rightarrow X$ given by $(g,x) \mapsto gx$, where
1. $ex=x$ for all $x \in X$ Identity
2. $(g_1g_2)x=g_1(g_2x)$ for all $x \in X$ and all $g_1,g_2 \in G$ Associative

Let $G=D_3$. If $X=\{1,2,3\}$ is a set, then when $D_3$ acts on $X$ we can see

$r \cdot 1=2$
$r \cdot 2=3$
$r \cdot 3=1$

$s \cdot 1=1$
$s \cdot 2=3$
$s \cdot 3=2$

Ex2: Let $G=D_4$, the symmetry group of a square. If $X=\{1,2,3,4\}$ is the set of vertices of the square, then we can consider $D_4$ to consist of the following permutations: $\{(1),(13),(24),(1432),(1234),(12)(34),(14)(23),(13)(24) \}$
The elements of $D_4$ act on $X$ as functions. The permutation (13)(24) acts on vertex 1 by sending it to vertex 3, on vertex 2 by sending it to vertex 4, and so on. It is easy to see that the axioms of a group action are satisfied.


$i \cdot H_1=\{i,-i,-1,1\}$
$j \cdot H_1=\{j,-j,-k,k\}$
$k \cdot H_1=\{j,-j,-k,k\}$


The orbit of a point x in X is the set of elements of X to which x can be moved by the elements of G. If $X$ is a $G$-set, then each partition of $x$ associated with $G$-equivalence is called an orbit of $X$ under $G$. We will denote the orbit that contains an element $x$ of $X$ by $O_x$.

Let $G=\{1,2,3,4,5,\dots, 26\}$ and $X=\{a,b,c,d,e,\dots, z\}$. Find $O_c$.

$O_c \cdot 1=d$
$O_c \cdot 2=e$
$O_c \cdot 3=f$ so on and so on.
therefore, $O_c=X$.
We see this because every letter can get hit by shifting.

Let $G=D_3$. If $X=\{1,2,3,4,5,6\}$, then find $O_3$. $hint:draw two triangles$
$O_3 \cdot 1=5$
$O_3 \cdot 2=1$
$O_3 \cdot 3=3$
therefore, $O_3=\{1,3,5\}$.

Also see $O_2=\{2,4,6\}$.


For every $x\in X$, we define the stabilizer subgroup of x (also called the isotropy subgroup) as the set of all elements in G that fix x:
Ex1 Let $G=D_3$ and $X=\{1,2,3\}$. If we think about the triangle, the stabilizer of the group would be the operation that keeps the triangle fixed in the same position. The fixed point $X$ under the action of $G$, $X_1=\{e,s\}=2$.

Orbit-Stabilizer Theorem

Theorem 12.3(Orbit-Stabilizer Theorem): Let $G$ be a finite group and $X$ a finite $G$-set. If $x \in X$, then $\left | O_x \right | = [G:G_x]=\left| G\right| /\left| G_x\right|$.

For every $x\in X$, we define the stabilizer subgroup of x (also called the isotropy subgroup) as the set of all elements in G that fix x:
$G_x=\{g\in G, g \times x=x \}$
$\left| Q_8 \right| =8$
$\left| G_H_1 \right| =4$
so, $\left|O_H_1 \right| =2$

Take G acting on X where $G=D_3$ and $X=\{1,2,3\}$, find $\left|0_x \right|$. We know from previous examples that $D_3$ has 2 stabilizers and $G$ has 6 elements, so by the Orbit-Stabilizer Theorem we see that $\left| 0_x\right| =\left| G\right|/ \left| G_x\right| =6/2=3$


Abstract Algebra: Theory and Applications

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License