# Groups Acting on Sets

we are going to analyze and classify groups, and, if possible, break down
complicated groups into simpler components. To motivate the topic of this blog, let’s
look at the following result.

## Cayley’s Theorem

Every group is isomorphic to a group of permutations.
Proof. The idea is that each element $g$ in the group $G$ corresponds to a permutation of the set $G$ itself. If $x \in G$, then the permutation associated with $g$ carries $x$ into $g x$. If $g x=g y$, then premultiplying by $g^{-1}$ gives $x=y$. Furthermore, given any $h \in G$, we can solve $g x=h$ for $x$. Thus the map $x \rightarrow g x$ is indeed a permutation of $G$. The map from $g$ to its associated permutation is injective, because if $g x=h x$ for all $x \in G$, then (take $x=1$ ) $g=h$. In fact the map is a homomorphism, since the permutation associated with $h g$ is multiplication by $h g$, which is multiplication by $g$ followed by multiplication by $h$, $h \circ g$ for short. Thus we have an embedding of $G$ into the group of all permutations of the set $G$
In Cayley’s theorem, a group acts on itself in the sense that each $g$ yields a permutation of $G$. We can generalize to the notion of a group acting on an arbitrary set.

The group $G$ acts on the set $X$ if for each $g \in G$ there is a mapping $x \rightarrow g x$ of $X$ into itself, such that

(1) $h(g x)=(h g) x$ for every $g, h \in G$
(2) $1 x=x$ for every $x \in X$.
As in $(5.1 .1), x \rightarrow g x$ defines a permutation of $X .$ The main point is that the action of $g$ is a permutation because it has an inverse, namely the action of $g^{-1}$. (Explicitly, the inverse of $x \rightarrow g x$ is $y \rightarrow g^{-1} y .$ ) Again as in (5.1.1), the map from $g$ to its associated permutation $\Phi(g)$ is a homomorphism of $G$ into the group $S_{X}$ of permutations of $X$. But we do not necessarily have an embedding. If $g x=h x$ for all $x$, then in (5.1.1) we were able to set $x=1$, the identity element of $G$, but this resource is not available in general. We have just seen that a group action induces a homomorphism from $G$ to $S_{X}$, and there is a converse assertion. If $\Phi$ is a homomorphism of $G$ to $S_{X}$, then there is a corresponding action, defined by $g x=\Phi(g) x, x \in X$. Condition (1) holds because $\Phi$ is a homomorphism, and (2) holds because $\Phi(1)$ must be the identity of $S_{X}$. The kernel of $\Phi$ is known as the kernel of the action; it is the set of all $g \in G$ such that $g x=x$ for all $x$, in other words, the set of $g$ ‘s that fix everything in $X$.

# The Orbit-Stabilizer Theorem

Suppose that the group $G$ acts on the set $X$. If we start with the element $x \in X$ and successively apply group elements in all possible ways, we get
$$B(x)=\{g x: g \in G\}$$
which is called the orbit of $x$ under the action of $G$. The action is transitive (we also say that $G$ acts transitively on $X$ ) if there is only one orbit, in other words, for any $x, y \in X$, there exists $g \in G$ such that $g x=y .$ Note that the orbits partition $X$, because they are the equivalence classes of the equivalence relation given by $y \sim x$ iff $y=g x$ for some $g \in G$. The stabilizer of an element $x \in X$ is
$$G(x)=\{g \in G: g x=x\},$$
the set of elements that leave $x$ fixed. A direct verification shows that $G(x)$ is a subgroup. This is a useful observation because any set that appears as a stabilizer in a group action is guaranteed to be a subgroup; we need not bother to check each time.

Before proceeding to the main theorem, let’s return to the examples considered in $(5.1 .3)$.

## Examples

1. The regular action of $G$ on $G$ is transitive, and the stabilizer of $x$ is the subgroup $\{1\}$.
2. The trivial action is not transitive (except in trivial cases), in fact, $B(x)=\{x\}$ for every $x$. The stabilizer of $x$ is the entire group $G$.
3. Conjugation on elements is not transitive (see Problem 1). The orbit of $x$ is the set of conjugates $g x g^{-1}$ of $x$, that is,
$$B(x)=\left\{g x g^{-1}: g \in G\right\},$$
which is known as the conjugacy class of $x$. The stabilizer of $x$ is
$$G(x)=\left\{g: g x g^{-1}=x\right\}=\{g: g x=x g\},$$
the set of group elements that commute with $x$. This set is called the centralizer of $x$, written $C_{G}(x)$. Similarly, the centralizer $C_{G}(S)$ of an arbitrary subset $S \subseteq G$ is defined as the set of elements of $G$ that commute with everything in $S$. (Here, we do need to check that $C_{G}(S)$ is a subgroup, and this follows because $\left.C_{G}(S)=\bigcap_{x \in S} C_{G}(x) .\right)$
4. Conjugation on subgroups is not transitive. The orbit of $H$ is $\left\{g H g^{-1}: g \in G\right\}$, the collection of conjugate subgroups of $H$. The stabilizer of $H$ is
$$\left\{g: g H g^{-1}=H\right\},$$
which is called the normalizer of $H$, written $N_{G}(H)$. If $K$ is a subgroup of $G$ containing $H$, we have
$H \unlhd K$ iff $g H g^{-1}=H$ for every $g \in K$

and this holds iff $K$ is a subgroup of $N_{G}(H)$. Thus $N_{G}(H)$ is the largest subgroup of $G$ in which $H$ is normal.
5. Conjugation on subsets is not transitive, and the orbit of the subset $S$ is $\left\{g S g^{-1}\right.$ :
$g \in G\} .$ The stabilizer of $S$ is the normalizer $N_{G}(S)=\left\{g: g S g^{-1}=S\right\}$.
6. Multiplication on left cosets is transitive; a solution of $g(x H)=y H$ for $x$ is $x=$ $g^{-1} y$. The stabilizer of $x H$ is
$$\{g: g x H=x H\}=\left\{g: x^{-1} g x \in H\right\}=\left\{g: g \in x H x^{-1}\right\}=x H x^{-1},$$
the conjugate of $H$ by $x$. Taking $x=1$, we see that the stabilizer of $H$ is $H$ itself.
7. Multiplication on subsets is not transitive. The stabilizer of $S$ is $\{g: g S=S\}$, the set of elements of $G$ that permute the elements of $S$.

## The Orbit-Stabilizer Theorem

Suppose that a group $G$ acts on a set $X .$ Let $B(x)$ be the orbit of $x \in X$, and let $G(x)$ be the stabilizer of $x$. Then the size of the orbit is the index of the stabilizer, that is,
$$|B(x)|=[G: G(x)] .$$
Thus if $G$ is finite, then $|B(x)|=|G| /|G(x)|$; in particular, the orbit size divides the order of the group.

Proof. If $y$ belongs to the orbit of $x$, say $y=g x$. We take $f(y)=g H$, where $H=G(x)$ is the stabilizer of $x$. To check that $f$ is a well-defined map of $B(x)$ to the set of left cosets of $H$, let $y=g_{1} x=g_{2} x$. Then $g_{2}^{-1} g_{1} x=x$, so $g_{2}^{-1} g_{1} \in H$, i.e., $g_{1} H=g_{2} H .$ Since $g$ is an arbitrary element of $G, f$ is surjective. If $g_{1} H=g_{2} H$, then $g_{2}^{-1} g_{1} \in H$, so that $g_{2}^{-1} g_{1} x=x$, and consequently $g_{1} x=g_{2} x$. Thus if $y_{1}=g_{1} x, y_{2}=g_{2} x$, and $f\left(y_{1}\right)=f\left(y_{2}\right)$
then $y_{1}=y_{2}$, proving $f$ injective.
Referring to $(5.2 .2)$, Example 3 , we see that $B(x)$ is an orbit of size 1 iff $x$ commutes with every $g \in G$, i.e., $x \in Z(G)$, the center of $G$. Thus if $G$ is finite and we select one element $x_{i}$ from each conjugacy class of size greater than 1 , we get the class equation
$$|G|=|Z(G)|+\sum_{i}\left[G: C_{G}\left(x_{i}\right)\right]$$
We know that a group $G$ acts on left cosets of a subgroup $K$ by multiplication. To prepare for the next result, we look at the action of a subgroup $H$ of $G$ on left cosets of $K$. Since $K$ is a left coset of $K$, it has an orbit given by $\{h K: h \in H\} .$ The union of the sets $h K$ is the set product $H K$. The stabilizer of $K$ is not $K$ itself, as in Example 6 ; it is $\{h \in H: h K=K\} .$ But $h K=K(=1 K)$ if and only if $h \in K$, so the stabilizer is $H \cap K$.

# Orbit-Counting Theorem

Let the finite group $G$ act on the finite set $X$, and let $f(g)$ be the number of elements of $X$ fixed by $g$, that is, the size of the set $\{x \in X: g(x)=x\}$. Then the number of orbits is
$$\frac{1}{|G|} \sum_{g \in G} f(g)$$
the average number of points left fixed by elements of $G$.
Proof. We use a standard combinatorial technique called “counting two ways”. Let $T$ be the set of all ordered pairs $(g, x)$ such that $g \in G, x \in X$, and $g x=x$. For any $x \in X$, the number of $g$ ‘s such that $(g, x) \in T$ is the size of the stabilizer subgroup $G(x)$, hence
$$|T|=\sum_{x \in X}|G(x)| .$$
Now for any $g \in G$, the number of $x$ ‘s such that $(g, x) \in T$ is $f(g)$, the number of fixed points of $g$. Thus
$$|T|=\sum_{g \in G} f(g)$$
Divide (1) and (2) by the order of $G$ to get
$$\sum_{x \in X} \frac{|G(x)|}{|G|}=\frac{1}{|G|} \sum_{g \in G} f(g)$$
But by the orbit-stabilizer theorem (5.2.3), $|G| /|G(x)|$ is $|B(x)|$, the size of the orbit of $x$. If, for example, an orbit has 5 members, then $1 / 5$ will appear 5 times in the sum on the left side of (3), for a total contribution of 1 . Thus the left side of (3) is the total number of orbits.
We can now proceed to the next step in the analysis.

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# MATH 355Abstract AlgebraFall 2021 (also offered Spring 2022)Division III Quantative/Formal Reasoning

CATALOG SEARCH

## Class Details

Algebra gives us tools to solve equations. The integers, the rationals, and the real numbers have special properties which make algebra work according to the circumstances. In this course, we generalize algebraic processes and the sets upon which they operate in order to better understand, theoretically, when equations can and cannot be solved. We define and study abstract algebraic structures such as groups, rings, and fields, as well as the concepts of factor group, quotient ring, homomorphism, isomorphism, and various types of field extensions. This course introduces students to abstract rigorous mathematics.The Class:Format: lecture
Limit: 30
Expected: 20
Class#: 1322
Grading: yes pass/fail option, yes fifth course option
Requirements/Evaluation:Problem sets and examsPrerequisites:MATH 250 or permission of instructorEnrollment Preferences:Students who have officially declared a major that requires Math 355.Distributions:Division IIIQuantative/Formal ReasoningQFR Notes:300-level math course