Group actions and some applications

Abstract algebra

Group theory

Group actions and some applications

Proposition 1

Suppose that G acts on X through an action $\star$ .Let $g\in G$ . Define a map $f_g$ from X to X by $f_g(x)=g\star x$ . Then, $f_g\in Sym(X)$.

Proof:

It is easy to verify.

Definition 1

Suppose that G acts on X through an action $\star$ . Define a map $\rho$ from G to $Sym(X)$ by $\rho(g)=f_g$. It is easy to verify that $\rho$ is a homomorphism from G to $Sym(X)$. Such a homomorphism is called a permutation representation of G on X

Remark

Given an action $\star$ of G on X,we have a representation of G on X. Conversely, given any representation $\rho$ of G on X, we have an action $\star$ of G on X defined by $g\star x=\rho(g)(x)$ such that the corresponding representation is the same as $\rho$. (It is easy to verify.)

Thus there is a faithful way to view an action as a representation , and a representation as an action , and they are related by

$$g\star x=\rho(g)(x)$$

Definition 2

Let $\star$ be an action of G on X with the corresponding representation $\rho$ . Then the kernel of $\rho$ is

$$ker \rho=\{g\in G|\rho(g)=1_G\}=\{g\in G|\rho(g)(x)=x,\forall x\in X\}=\{g\in G|g\star x=x,\forall x\in X\}$$

Therefore , $\{g\in G|g\star x=x,\forall x\in X\}$ is a normal subgroup of G. This subgroup is called isotropy group or the stabilizer of the action $ \star$ of G on X , and it is denoted by $Stab(G,X)$

By the fundamental theorem of homomorphism , we have the following proposition.

Proposition 2

$G/Stab(G,X)$ is isomorphic to the subgroup $\rho(G)$ of $Sym(X)$ , where $\rho$ is the corresponding representation.

Remark

Suppose that G acts on X through an action $\star$​ .Then we have

$$Stab(G,X)=\bigcap\limits_{x\in X}Stab(x)$$

where $Stab(x)=\{g\in G|g\star x=x\}$ is the stabilizer of $x\in X$.

Definition 3

Let G be a group which acts on X through an action $\star$ . An element $x\in X$ is called a fixed point if $g\star x=x$ , for all $g\in G$ . Thus , x is a fixed point if and only if $Stab(x)=G$ , or equivalently ,$Orb(x)=\{x\}$. The set of all fixed points of the action $\star$ is denoted by $X^G$ ,and it is called the fixed point set of the action. Thus $X^G=\{x\in X|g\star x=x,\forall g\in G\}$.

Remark

An element $x\in X$ is a fixed point if and only if the orbit $Orb(x)$ is singleton . Thus $X^G$ is the union of all those orbits which are singletons.

Proposition 3

Let A be a set obtained by choosing one and only one member from each orbit different from singleton. Suppose that X is finite , then we have the equation

$$|X|=|X^G|+\sum\limits_{x\in A}[G:Stab(x)]$$

Here again ,each term under summation in $R.H.S. $​ is greater than 1 and divides |G|

The equation is also called the class equation or class formula for the action $\star$.

Proof:

If $x\in A$ ,then $Orb(x)\neq \{x\}$. Further, $Orb(x)\cap Orb(y)=\varnothing$ for all $x,y\in A,x\neq y$.

Since X is union of all orbits,

$$X=X^G\bigcup(\bigcup\limits_{x\in A}Orb(x))$$

Since distinct orbits are disjoint and $|Orb(x)|=[G:Stab(x)]$ , we have

$$|X|=|X^G|+\sum\limits_{x\in A}[G:Stab(x)]$$

Now, we give some examples and applications to the structure theory of finite groups.

Example 1

Let G be a group and X=G . Define $g\star x=gxg^{-1},g,x\in G$. It is easy to verify that $\star$ is an action of G on G. This action is called the inner conjugation of G on G.

The isotropy group of the inner conjugation of G on G at x is $Stab(x)=\{g\in G|gxg^{-1}=x\}=\{g\in G|gx=xg\}=C_{(x)}$ , in which $C_{(x)}$ is the centralizer of x in G.

The isotropy group of the inner conjugation of G on G is $ker\rho =Stab(G,X)=\{g\in G|gxg^{-1}=x,\forall x\in G\}=\{g\in G|gx=xg\}=C$ , in which $\rho$ is the corresponding representation , C is the center of G. The representation $\rho$ determined by the inner conjugation of G on G is given by $\rho(g)=f_g$ , where $f_g$ is the inner automorphism of G determined by g. Thus , $\rho(G)$ is the group $Inn(G)$​ of all inner automorphisms of G. By the fundamental theorem of homomorphism ,we have the following:

$$G/C\cong Inn(G)$$

The orbit through $x\in G$​ of the inner conjugation of G on G is the set $Orb(x)=\{gxg^{-1}|g\in G\}$​​​ of all conjugates of x in G ,which is also called the conjugacy class of G determined by x.

The fixed point set of the inner conjugation of G on G is the set $G^G=\{x\in G|gxg^{-1}=x,\forall g\in G\}=C$ .

Let A be a set obtained by choosing one and only member from each orbit different from singleton . By the class formula,we have the following:

$$|G|=|C|+\sum\limits_{x\in A}[G:Stab(x)]$$

This equation is also called the classical class equation of G.

Example 2

Let G be a group and X the set S(G) of all subgroups of G . Define $g\star H=gHg^{-1},g\in G$. It is easy to verify that $\star$ is an action of G on S(G). This action is called the inner conjugation of G on subgroups of G.

The isotropy group of the inner conjugation of G on subgroups of G at H is $Stab(H)=\{g\in G|gHg^{-1}=H\}=\{g\in G|gH=xH\}=N(H)$ , in which $N(H)$ is the normalizer of x in G.

The orbit through $H\in S(G)$ of the inner conjugation of G on subgroups of G is the set $Orb(H)=\{gHg^{-1}|g\in G\}$ of all conjugates of H in G ,which is also called the conjugacy class of G determined by H.

The fixed point set of the inner conjugation of G on subgroups of G is the set $S(G)^G=\{H\in S(G)|gHg^{-1}=H,\forall g\in G\}=\{H\in S(G)|gH=Hg,\forall g\in G\}=\{H\in S(G)|H\lhd G\}$ ,which is the set of all normal subgroups of G.

Let A be a set obtained by choosing one and only member from each orbit different from singleton . By the class formula,we have the following:

$$|S(G)|=|S(G)^G|+\sum\limits_{x\in A}[G:N(H)]$$

Example 3

Let G be a group and $X=G/^l H$ . Define $g\star xH=gxH,g,x\in G$. It is easy to verify that $\star$ is an action of G on $G/^lH$. This action is called the left multiplication of G on $G/^lH$.

The isotropy group of the left multiplication of G on $G/^lH$ at $xH$ is $Stab(xH)=\{g\in G|gxH=xH\}=\{g\in G|x^{-1}gx\in H\}=xHx^{-1}$

The isotropy group of the left multiplication of G on $G/^lH$ is $ker\rho =Stab(G,G/^lH)=\bigcap\limits_{x\in G}Stab(xH)=\bigcap\limits_{x\in G}xHx^{-1}$ , in which $\rho$ is the corresponding representation . Thus , $\bigcap\limits_{x\in G}xHx^{-1}$ is a normal subgroup of G . Further , it is the largest normal subgroup of G contained in H. This subgroup is called the core of H in G ,and it is denoted by $Core_G(H)$ . By the fundamental theorem of homomorphism , $G/Core_G(H)$ is isomorphic to a subgroup of $Sym(X)$.

Now we apply the theory of group actions to the structure theory of finite groups.

Proposition 4

Let G be a group of order $p^n,n\ge 1$ ,where p is a prime. Let H be a nontrivial subgroup of G. Then $H\cap C(G)\neq \{e\} $ . In particular, $C(G)\neq \{e\}$.

Proof:

Since $H\lhd G$ , G acts on H through inner conjugation . The fixed point set

$$H^G=\{h\in H|ghg^{-1}=h,\forall g\in G\}=\{h\in H|gh=hg,\forall g\in G\}=H\cap C(G)$$

Let A be a set obtained by choosing one and only member from each orbit different from singleton , consider the class formula ,

$$p^r=|H|=|H\cap C(G)|+\sum\limits_{x\in A}[G:Stab(x)]$$

Each term under summation in $R.H.S. $ is greater than 1 and divides |G|

Since $H\neq \{e\},r\ge 1$ .Thus , $L.H.S.$ is divisible by p and the second term in the $R.H.S.$ is also a multiple of p . Since p divides $|H\cap C(G)|$ . Since $ e\in H\cap C (G)$ . $H\cap C(G)$ contains at least p elements.

Taking H=G , we observe that $C(G)\neq \{e\}$.

Corollary

Let G be a group of order $p^2$ , where p is a prime . Then, G is abelian.

Proof:

Consider $C(G)$ . By the above proposition ,|C(G)|=p or $p^2$ . If $|C(G)|=p^2$ , then C(G)=G ,and so G is abelian. Now we show that $|C(G)|\neq p$ .

Suppose that$|C(G)|= p$ , then $|G/C(G)|=p$ ,and so G is cyclic . Thus G is abelian ,and then G=C(G) . So $|C(G)|=|G|= p^2$ . This is a contradiction to the supposition.

Proposition 5

Let G be a group of order $p^n,n\ge 1$ ,where p is a prime. Let H be a subgroup of G and $H\neq G$ . Then, $ H\neq N(H)$​ , where N(H) is the normalizer of H in G.

Proof:

The proof is by induction on $n$ .

If $n=1$ ,then G is prime cyclic . And then $H\neq G=N(H)$.

Assume that the result is true for a group of order less than $p^n$. Let G be a group of order $p^{n}$ .

If $C(G)\nsubseteq H$ ，then there is an element $x\in C(G)$ but $x\notin H$ . Since $xH=Hx,x\in N(H)$. Thus,$ H\neq N(H)$ . ​

If $C(G)\subseteq H$ , consider the group $G/C(G)$ . By proposition 4 , $C(G)\neq \{e\}$ ,and then $|G/C(G)|<p^{n}$ . By the induction hypothesis , for all subgroup $H/C(G)$ of $G/C(G), H\neq G$ ,we have $N(H/C(G))\neq H/C(G)$ . Then there is an element $x\in G-H$ ,such that $\overline{xHx^{-1}}=\overline H$ . By the correspondence theorem, we have $xHx^{-1}=H$ ,which implies that $x\in N(H), and,H\neq N(H)$.

Corollary

Let G be a group of order $p^n,n\ge 1$ ,where p is a prime. Let H be a subgroup of G of order $p^{n-1}$. Then , H is a normal subgroup of G.

Proof:

By the above proposition, $H\subset N(H)$ and $H\neq N(H)$ . Since $|H|=p^{n-1}$ and N(H) is also a subgroup of G whose order is greater than $p^{n-1}$ and divides $p^n$. Thus,$|N(H)|=p^n$ . This shows that N(H)=G and $H\lhd G$.

Proposition 6

Let G be a finite group and p the smallest prime dividing the order of G. Then, every subgroup of G of index p is a normal in G.

Proof:

Let H be a subgroup of G of index p, where p is the smallest prime dividing the order of G. Then $G/^l H$ contains p elements. Consider the action $\star$ of G on $G/^l H$ defined by $g\star xH=gxH$ . The isotropy group of this action is the $Core_G(H)$, which is the largest normal subgroup of G contained in H. Also, $G/Core_G(H)$ is isomorphic to a subgroup of $Sym(G/^lH)\cong S_p$ . Since $|G/Core_G(H)|=[G:H][H:Core_G(H)]=pq$ . We have

$$pq|p!\Rightarrow q|(p-1)\cdots 1$$

Since p is the smallest prime dividing the order of G and q divides |G|, q=1 . Thus $H=Core_G(H)\lhd G$

Corollary

If G have no subgroups of index 2 . Then ,every subgroup of index 3 must be normal in G.

Proof:

Let H be a subgroup of G of index 3. Consider $Core_G(H)$ .By the fundamental theorem of homomorphism, $G/Core_G(H)$ is isomorphic to a subgroup of $Sym(G/^lH)\cong S_3$ . If $G/Core_G(H)$ is isomorphic to $S_3$ , since $S_3$ have a subgroup of index 2 , G must have a subgroup of index 2 (by the correspondence theorem), which contradiction to the hypothesis. Since $|G/Core_G(H)|=[G:H][H:Core_G(H)]\ge 3$ , then $|G/Core_G(H)|=3$ ,which implies that $H=Core_G(H)\lhd G$.

PermaLink:
https://tuowang2002.github.io/2021/09/10/group11/