LEC ALG2 8

Orbit, stabilizer and kernel

Definition 97.1(Definition).

Let $\varphi :G\times X\to X$ be a group action. For $x\in X$, define the orbit of $x$ by

$${\theta }_x\equiv \{gx|g\in G\}.$$

The action of $G$ on $X$ is called transitive if there is only one orbit.

Note that ${\theta }_x\subset X$, and does not have any group structure.

Definition 97.2(Definition).

Let $\varphi :G\times X\to X$ be a group action. For $x\in X$, define the stabilizer of $x$ by

$$G_x\equiv \{g\in G|gx=x\}.$$

It is easy to verify that $G_x<G$ for all $x\in X$.

Definition 97.3(Definition).

The kernel of a group action $\varphi$ is the set of all elements $g\in G$ for which ${\varphi }_g$ is the identity on $S$:

$$\ker \varphi =\{g\in G|{\varphi }_g(s)=s\forall s\in S\}.$$

Note that $\ker \varphi ={\bigcap }_{x\in X}{G}_x◃G$.

Example 97.4(Example).

In Exm 96.5, ${\theta }_{{\Pi }_1}={\theta }_{{\Pi }_2}={\theta }_{{\Pi }_3}=X$, and $G_{{\Pi }_1}=\{e,(12)(34),(13)(24),(14)(23),(1324),(1423),(12),(34)\}$, which is just the union of the inverse images of $({\Pi }_1,{\Pi }_2,{\Pi }_3)$ and $({\Pi }_1,{\Pi }_3,{\Pi }_2)$. Ditto for $G_{{\Pi }_2}$ and $G_{{\Pi }_3}$. The kernel is $\{e,(12)(34),(13)(24),(14)(23)\}$.

Example 97.5(Example).

Let $G=G{L}_2(\mathbb{R})$ and $X={\mathbb{R}}^2$. Let $\varphi :G\times X\to X$ be defined by $\varphi (M,\mathbf{v})=M\mathbf{v}$, $i$.$e$, $G$ acts on $X$ by left multiplication (verify that this is a group action).

For $\mathbf{x}=\left[\begin{array}{c}0\\ 0\end{array}\right]$, ${\theta }_{\mathbf{x}}=\{\mathbf{x}\}$, $G_{\mathbf{x}}=G$.
For $\mathbf{x}=\left[\begin{array}{c}1\\ 0\end{array}\right]$, ${\theta }_{\mathbf{x}}={\mathbb{R}}^2-\{0\}$, $G_{\mathbf{x}}=\left[\begin{array}{cc}1& \ast \\ 0& d\ne 0\end{array}\right]$.

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Groups acting on themselves

Let $G$ be a group, and $X=G$. Standard group actions include

1. Left multiplication: $\varphi (g,x)=gx\equiv g\circ x$.
2. Right multiplication: $\varphi (g,x)=gx\equiv x\circ g^{-1}$.
3. Conjugation: $\varphi (g,x)=gx\equiv g\circ x\circ g^{-1}$. The stabilizer of $x$ under conjugation, also denoted by $Z(x)$, is called the centralizer of $x$. It is the set of all $g\in G$ which commute with $x$. Note that the orbit of $x$, $\{{\varphi }_g(x)|g\in G\}$, is the conjugacy class of $X$. The centralizer of any subset $A\subset G$ is similarly defined.

$G$ can also act on $G/H$ for some $H<G$: ${\varphi }_g(aH)=(ga)H$.

Theorem 97.6(Theorem).

Let $G$ be a group, let $H<G$, let $G$ act by left multiplication on $G/H$. Let ${\pi }_H$ be the permutation representation. Then,

1. this action is transitive.

2. the stabilizer of $aH\in G/H$ is $G_{aH}=aH{a}^{-1}$.

3. the kernel of the action is ${\bigcap }_{x\in G}xH{x}^{-1}$, and $\ker {\pi }_H$ is the largest normal subgroup of $G$ contained in $H$.

Proof.

To see (1), observe that for any two cosets$aH,bH\in G/H$, $(b{a}^{-1})aH=bH$. Thus, any two arbitrary elements of $G/H$ lie in the same orbit.

For (2), we have

$$\begin{array}{rl}{G}_{aH}& =\{g\in G|g(aH)=aH\}\\ & =\{g\in G|(ga)H=aH\}\\ & =\{g\in G|a^{-1}ga\in H\}\\ & =\{g\in G|g\in aH{a}^{-1}\}\\ & =aH{a}^{-1}.\end{array}$$

For (3), we already know that the kernel of a group action is the intersection of all stabilizers, so $\ker {\pi }_H={\bigcap }_{x\in G}xH{x}^{-1}$. We know that $\ker {\pi }_H◃G$. Further, if $g\notin H$, then $g(1H)=gH\ne 1H$, so $g\notin \ker {\pi }_H$. Thus, $\ker {\pi }_H\le H$. Let $N$ be a normal subgroup of $G$ contained in $H$. For every $x\in G$, conjugation by $x$ is a homomorphism, so $N\le H$ implies $xN{x}^{-1}\le xH{x}^{-1}$. But $xN{x}^{-1}=N$, so we have $N\le xH{x}^{-1}$ for all $x\in G$. This implies

$$N\le \bigcap _{x\in G}xH{x}^{-1}=\ker {\pi }_H.$$

Thus, $\ker {\pi }_H$ is the largest normal subgroup of $G$ contained in $H$.□

Corollary 97.7(Corollary).

Let $p$ be the smallest prime dividing the order of a group $G$. Then, any subgroup of $G$ of index $p$ is a normal subgroup.

Proof.

Suppose$H\le G$ and $[G:H]=p$. Let ${\pi }_H$ be the permutation representation of $G$ acting on the set of all left cosets of $H$ in $G$ by left multiplication (just to be clear, ${\pi }_H:G\to S_{G/H}$). Let $K=\ker {\pi }_H$. Then, $K\le H$ and $K◃G$. Let $[H:K]=k$. Then, $[G:K]=pk$.

From the first isomorphism theorem, $G/K\cong \mathrm{Im}{\pi }_H$. Since $\mathrm{Im}{\pi }_H\le S_{G/H}\cong S_p$, $|\mathrm{Im}{\pi }_H|$ divides $p!$. So, $|G/K|=pk$ divides $p!$. Thus $k|(p-1)!$. But, since $K$ is a subgroup of $G$, $k||G|$, and so all prime divisors of $k$ are greater than or equal to $p$, forcing $k$ to be $1$. So, $H=K◃G$.□

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Centralizer and normalizer

Definition 97.8(Definition).

The centralizer of $A\subseteq G$ is the set of elements of $G$ which commute with every element of $A$. Denoted by $Z(A)$ or $C_G(A)$.

It can easily be shown that $C_G(A)<G$ for all $A\subseteq G$.

Definition 97.9(Definition).

The center of a group $G$ is defined to be $C_G(G)$, the set of all elements in $G$ which commute with every element of $G$. Also denoted by $Z(G)$.

It follows that $Z(G)<G$.

Definition 97.10(Definition).

The normalizer of $A$ in $G$ is the set of all $g\in G$ that satisfy $gA=Ag$: $N_G(A)=\{g\in G|gA{g}^{-1}=A\}$.

It can be shown that $N_G(A)<G$ for all $A\subseteq G$.

More details here. Refer Dummit & Foote (2004, p. 50) for examples.

Note the crucial difference between the centralizer and the normalizer:

$$\begin{array}{rl}& C_G(A)=\{g\in G|ga{g}^{-1}=a\forall a\in A\},\\ & N_G(A)=\{g\in G|gA{g}^{-1}=A\}.\end{array}$$

It follows that $C_G(A)\subseteq N_G(A)$, and since $C_G(A)$ is a group, we have $C_G(A)\le N_G(A)$.

In summary, we have:

Important

$Z(G)\le C_G(A)\le N_G(A)\le G$ for all $A\subseteq G$.

How these arise naturally from group actions

Let $G$ be a group. Let $\mathcal{P}(G)$ be the power set of $G$. Let $G$ act on $\mathcal{P}(G)$ by conjugation:

$$A\stackrel{{\varphi }_g}{\mapsto }gA{g}^{-1},A\in \mathcal{P}(G),g\in G.$$

Let $A\subseteq G$. Under this action, the normalizer of $A$ in $G$ is the stabilizer of $A$: $N_G(A)=G_A$.

Next, let the group $N_G(A)$ act on $A$ by conjugation:

$$a\stackrel{{\psi }_k}{\mapsto }ka{k}^{-1},a\in A,k\in N_G(A).$$

Under this action, the centralizer of $A$ in $G$ is the kernel: $C_G(A)=\ker \psi$.

Next, let the group $C_G(A)$ act on $G$ by conjugation:

$$g\stackrel{{\chi }_h}{\mapsto }hg{h}^{-1},g\in G,h\in C_G(A).$$

Under this action, the center of $G$ is the kernel: $Z(G)=\ker \chi$. So, we have $Z(G)\le C_G(A)\le N_G(A)\le G$.

Some theorems

Theorem 97.11(Theorem).

If $G/Z(G)$ is cyclic, then $G$ is abelian.

Proof.

Let$G/Z(G)=⟨aZ(G)⟩$ for some $a\in G$. Then, any $g\in G$ can be expressed as $a^{k}z$ for some integer $k$ and $z\in Z(G)$. Let $g_1,g_2\in G$, $g_1=a^{k_1}{z}_1$, $g_2=a^{k_2}{z}_2$. Then, $g_1{g}_2=a^{k_1}{z}_1{a}^{k_2}{z}_2=a^{k_1+k_2}{z}_1{z}_2$, and $g_2{g}_1=a^{k_2+k_1}{z}_2{z}_1=a^{k_1+k_2}{z}_1{z}_2$. Therefore, $G$ is abelian.□

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Second isomorphism theorem

Theorem 97.12(Theorem).

Let $G$ be a group, let $A$ and $B$ be subgroups of $G$ and assume $A\le N_G(B)$. Then,

1. $AB\le G$,

2. $B⊴AB$,

3. $A\cap B⊴A$, and

4. $AB/B\cong A/A\cap B$.

Proof.

$(1)$ follows from here. Since $A\le N_G(B)$ and $B\le N_G(B)$, $AB\le N_G(B)$, so $B⊴AB$. This allows us to define the homomorphism

$$\varphi :A\to AB/B,a\mapsto aB.$$

It is easy to see that $\varphi$ is a homomorphism: $\varphi (ab)=abB=(aB)(bB)=\varphi (a)\varphi (b)$. Clearly, $\ker \varphi =A\cap B$. Thus, $A\cap B⊴A$. Additionally, since $\varphi$ is surjective, the first isomorphism theorem gives $AB/B\cong A/A\cap B$.□

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References

Dummit, D. S., & Foote, R. M. (2004). Abstract Algebra (3rd ed). Wiley.