Source of most content: Gunhee Cho

Group action

Def. Let $G$ be a group, $X$ be a set. We call thefunction $\cdot : G \times X \rightarrow X \ ((g, x) \mapsto g\cdot x$ the group action of $G$ on $X$ if it satisfies:

1. (Identitiy) $\forall x \in X, \ e \cdot x = x$.
2. (Compatibility) $\forall g_1, g_2 \in G, \forall x \in X, \ (g_1g_2)\cdot x = g_1 \cdot ( g_2 \cdot x)$, i.e., group action is compatible with group operation.
   We call X a $G$-set and G acts on X which is denoted by $G \curvearrowright X$.

e.g.) (conjugate action) $G$ any group, $X = G$, and $\forall g \in G, \forall x \in X = G, \ g\cdot x := gxg^{-1}$, Then $G \curvearrowright X$.
e.g.) $G$ any group, $X=G$, if $g \cdot x := gx$ (left multiplication) then $G \curvearrowright X$ by left multiplication.
e.g.) $X = \{ x\}, G$ any group, $\forall g \in G, \ g \cdot x = x$. This $G$-set $X$ is called a trivial $G$-set.
e.g.) $H \leqslant G, X = G/H := \{ xH \mid x \in G \}$, a collection of left coset, $\forall g \in G, \forall x \in H \in G/H, g \cdot xH := gxH$, where $gx$ is group operation. Therefore, $G \curvearrowright X$ by left multiplication.

Group action and homomorphism

Let $G \curvearrowright X$ with $\cdot$. For each $g\in G$, define $\varphi_g : X \rightarrow X, \ \varphi_g(x) = g \cdot x, \ \forall x \in X$. Then, $\varphi_g$ is bijective.

1. (injective) If $\varphi_g(x_1) = g \cdot x_1 = \varphi_g(x_2) = g \cdot x_2 \Leftrightarrow g^{-1}\cdot (g \cdot x_1)=g^{-1}\cdot (g \cdot x_2) \leadsto x_1=x_2$.
2. (surjective) Take any $y \in X$ and any $x=g^{-1}\cdot y$. Then, $\varphi_g(x)=g\cdot (g^{-1}\cdot y)= (g\cdot g^{-1)}\cdot y=e \cdot y = y$.
   

Then, we can define $\rho:G \rightarrow S_X \ (g \mapsto \varphi_g)$, where $S_X$ is a symmetric group of $X$ which consists of bijective functions from $X$ to itself and is a group w.r.t. the function composition. Then, $\rho(gg’)=\rho(g)\cdot \rho(g’) \Leftrightarrow \varphi_{gg’}=\varphi_{g}\cdot\varphi_{g’}$, since for each $x\in X, \ \varphi_{gg’}(x)=gg’ \cdot x = g\cdot (g’ \cdot x) = g \cdot \varphi_{g’}(x) = \varphi_g (\varphi_{g’}(x))$. Therefore, $\rho$ is group homomorphism.

Conversely, given a $G$-set $X$, we can define a group action of $G$ on $X$ if we have a group homomorphism $\tilde{\rho}: G \rightarrow S_X$ as $\forall g \in G, \ \forall x \in X, g\cdot x := \tilde{\rho}(g)(x) \in X$, where $\tilde{\rho}$ is a bijective function from $X$ to $X$. Then, $\cdot$ is a group action since $e \cdot x = \tilde{\rho}(e)(x)=x$ ($\tilde{\rho}(e)=e$ since $\tilde{\rho}$ is gp homomorphism) and $(g_1g_2)\cdot x = \tilde{\rho}(g_1 g_2)(x) =(\tilde{\rho}(g_1) \cdot \tilde{\rho}(g_2))(x)$$ = g_1 \cdot (g_2 \cdot x)$.

Def. For a set $X$, a group homomorphism $\rho: G \rightarrow S_X$ is called a permutation representation of $X$.

Cayley’s theorem

Any group $G \curvearrowright G$ by left multiplication, i.e., $X=G$ is a $G$-set, then we have group homomorphism $\lambda: G \rightarrow S_G \ (g \mapsto \lambda(g))$, where $\lambda(g): G \rightarrow G \ (h \mapsto \lambda(g)(h)=g\cdot h = gh)$.

\[\text{Then, this } \lambda \text{ is injective, i.e., } G \cong \lambda(G) \leqslant S_G. \nonumber\]

$\implies$ Any group is a subgroup of a symmetric group.

Proof

It suffices to show that $\ker \lambda = \{ e_G \}$ since $\lambda$ is gp homomorphism (check proposition). Let $x \in \ker \lambda$. Then, by definition, $\lambda(x) = Id_G$ (identity element of $S_G$) $\leadsto \lambda(x)(e_G) = e_G \leadsto x \cdot e_G =e_G \Leftrightarrow x=e_G$, which implies that $\ker \lambda$ only contain $e_G$ as an element.

Orbit and subgroup

Def. For a $G$-set $X$, $\forall x \in X, \ G \cdot x = \{g \cdot x \mid g \in G \}$ is called $G$-orbit of $x$ and $G_x := \{ g\in G \mid g \cdot x = x \}$ is called isotropy subgroup for $x$.

NB1; For $(G, *)$, $G_x \leqslant G$ since (1) $G_x \subseteq G$, (2) $\forall g_1, g_2 \in G_x, \ (g_1 * g_2)\cdot x = g_1 \cdot (g_2\cdot x) =g_1 \cdot x =x$$ \leadsto g_1 * g_2 \in G_x$ and (3) $\forall g \in G_x, \ g^{-1} * g = e_G, \ (g^{-1} * g) \cdot x = e \cdot x = x = g^{-1} \cdot (g \cdot x) = g^{-1} \cdot x \leadsto g^{-1} \in G_x$.
NB2; $\forall x \in X, \ \varphi_x: G/G_x \rightarrow G \cdot x \ (gG_x \mapsto g \cdot x)$ is a bijective function.

1. (injective) If for $g_1, g_2 \in G, \ h_1, h_2 \in G_x, \varphi_x (g_1h_1) = \varphi_x (g_2h_2) \leadsto g_1 \cdot x = g_2 \cdot x$. By definition of $G_x$, $h_1 \cdot x = h_2 \cdot x = x \leadsto g_1 \cdot (h_1 \cdot x) = g_2 \cdot (h_2 \cdot x) = (g_1 * h_1)\cdot x = (g_2 * h_2)\cdot x, \forall x \in X$. Thus $g_1 h_1 = g_2 h_2$.
2. (surjective) For any $g_0 \in G, \ \forall x \in X, \ \varphi_x (g_0 G_x) = g_0 \cdot x \implies \text{Im}\varphi_x = G \cdot x$.

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\[\text{Let } H \leqslant G, \text{ then group operation in } G/H:=\{ gH \mid g \in G \} \text{ is well-defined iff } yxy^{-1}\in H, \forall y\in G, \forall x \in H. \nonumber\]

Here, $G/H$ is a set of left cosets of $H$ in $G$.

Proof

1. $(\Rightarrow)$ Let $xH=\{xh \mid h \in H \} = H$, then $x \in xH \leadsto x \in H$. Since we assume group operation is well-defined, $xH * yH = xyH \leadsto xH * yH = H * yH = yH = xyH \leadsto xyH = yH \Leftrightarrow y^{-1}xyH=H \Leftrightarrow y^{-1}xy \in H$.
2. $(\Leftarrow)$ Let $xH=x’H,\ yH=y’H \Leftrightarrow x(x’)^{-1} \in H, y(y’)^{-1} \in H$. Then, we need to show that $xyH=x’y’H$, i.e., group operation is well-defined. Since $(xy)(x’y’)^{-1} = xy(y’)^{-1}(x’)^{-1}=x * y(y’)^{-1} * x^{-1} * x(x’)^{-1} \in H$ since $x, y(y’)^{-1}, x(x’)^{-1}, x^{-1} \in H$, group operation $*$ is well-defined.

Normal subgroup

Def. We call $H\leqslant G$ a normal subgroup if $\forall g \in G, \forall h \in H, \ ghg^{-1}\in H$, which is denoted by $H \trianglelefteq G$.

NB3; $H \trianglelefteq G \Leftrightarrow \forall g \in G, \ gHg^{-1}=H \Leftrightarrow \forall g \in G, \ gH=Hg$.

Proposition/theorem

\[\text{Let } \Phi: G \rightarrow G' \text{ be a group homomorphism. Then, } \ker \Phi \trianglelefteq G. \nonumber\]

Proof

We need to show that $\forall g \in G, \forall h \in \ker \Phi, \ ghg^{-1} \in \ker \Phi$. $\Phi(ghg^{-1}) \stackrel{(a)}{=} \Phi(g) \Phi(h) \Phi(g^{-1}) \stackrel{(b)}{=} \Phi(g)\Phi(g^{-1}) \stackrel{(a)}{=} \Phi(g) (\Phi(g))^{-1}=e$, where (a) holds because $\Phi$ is a group homomorphism and (b) holds since $h \in \ker \Phi$.

NB4; For a given group homomorphism $f:G \rightarrow H$, we showed that $\ker f \leqslant G, \text{Im}f \leqslant H$ in Propositon. We also show that $\ker f \trianglelefteq G$. However, in general, $\text{Im}f \trianglelefteq H$ does not hold.

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\[\text{Let } f: H \rightarrow G \text{ be a group homomorphism. Assume } \ker f \subseteq H \leqslant G, \text{ then } f^{-1}(f(H))=H. \nonumber\]

Proof

1. $(\supseteq)$ $f^{-1}(f(H)) \supseteq H$ in general. Check
2. $(\subseteq)$ Let $x \in f^{-1}f(H) := \{ y \mid f(y) \in f(H)\} \Leftrightarrow f(x) \in f(H)$. This implies that $\exists h \in H$ s.t. $f(x)=f(h) \Leftrightarrow f(x)f(h)^{-1}=f(xh^{-1})=e \leadsto xh^{-1}\in \ker f \subseteq H \leadsto \exists h’ \in H$ s.t. $xh^{-1}=h’ \leadsto x = hh’ \in H$. Therefore, if $x \in f^{-1}(f(H))$ then $x \in H$.

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\(\text{Let } \phi: G \rightarrow G' \text{ a group epimorphism. Consider } \mathcal{A}:= \{ H \subseteq G \mid \ker \phi \subseteq H \leqslant G \}, \ \mathcal{B}:= \{ H' \subseteq G' \mid H' \leqslant G' \}. \nonumber\) \(\text{Define } \Phi: \mathcal{A} \rightarrow \mathcal{B} \ (H \mapsto \phi(H)). \text{ Then, } \Phi \text{ is bijective, i.e., if } H' \in \mathcal{B}, \text{ then, } \Phi^{-1}(H')=\phi^{-1}(H'). \nonumber\)

Proof

1. $\forall H \leqslant G, \phi(H) \leqslant G’$ since $\phi$ is a group homomorphism. Therefore, $\Phi$ is well-defined.
2. (injective) Let $\Phi(H_1) = \Phi(H_2) \Leftrightarrow \phi(H_1) = \phi(H_2)$. By above theorem, $\phi^{-1}(\phi(H_1)) = H_1, \ \phi^{-1}(\phi(H_2)) = H_2$. Therefore, $H_1 = H_2$.
3. (surjective) Take any $H’ \leqslant G’$. Choose $H_0 = \phi^{-1}(H’)$ (inverse image of $H’$). Since $\phi^{-1}(H’) \leqslant G$ and $\ker \phi \subseteq \phi^{-1}(H’)$ (if $x\in \ker \phi, \phi(x)=e_{G’} \in H’$), $H_0 \in \mathcal{A}$. Therefore, by above theorem, $\phi^{-1}(\phi(H_0))=H_0$. Since $\phi$ is surjective, $\phi(\phi^{-1}(H’))=H’$. Thus, $\Phi(H_0) := \phi(H_0)= H’$.
   $\implies$ $\Phi^{-1}$ is well-defined and $\Phi^{-1}(H’)=H_0 = \phi^{-1}(H’)$.

NB5; Let $\phi: G \rightarrow G’$ be a group epimorphism and $\tilde{\mathcal{A}}:= \{ H \subseteq G \mid \ker \phi \subseteq H \trianglelefteq G \},$ $\tilde{\mathcal{B}}:= \{ H’ \subseteq G’ \mid H’ \trianglelefteq G’ \}$. Define $\Phi: \tilde{\mathcal{A}} \rightarrow \tilde{\mathcal{B}} \ (H \mapsto \phi(H))$. Since $\ker \phi \subseteq H \trianglelefteq G \leadsto \phi(H) (=: \Phi(H)) \trianglelefteq \phi(G) = G$ (since $\phi$ is surjective), $\Phi$ is well-defined. Similarly, we can show that $\Phi$ is bijective, i.e., there is one-to-one correspondence between $\{ H \subseteq G \mid \ker \phi \subseteq H \trianglelefteq G \} \leftrightarrow \{ H’ \subseteq G’ \mid H’ \trianglelefteq G’ \}$ if $\phi: G \rightarrow G’$ is a group epimorphism.

Maximal normal subgroup

Def. A group $G$ is simple if there is no normal subgroup except $\{e\}$ and $G$.

Def. $M \trianglelefteq G$ is a maximal normal subgroup $M$ of $G$ if $M\ne G$ and $\exists N \subseteq G$ s.t. if $M \subsetneq N \trianglelefteq G$ then $N=G$.

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\(\text{Let } N \trianglelefteq G, \ \pi:G \rightarrow G/N \ (g \mapsto gN) \text{ be a group epimorphism, where we have }\ker \pi = \{g \in G \mid gN=N \}=N. \nonumber\) \(\text{Then, } \Phi: \{ H \subseteq G \mid \ker \pi = N \trianglelefteq G \} \rightarrow \{ H' \subseteq G/N \mid H' \trianglelefteq G/N \} \text{ is bijective.} \nonumber\)

Proof follows from NB5 and this implies that there is one-to-one correspondence between (normal) subgroup of $G$ containing $N$ and (normal) subgroup of $G/N$.

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\[\text{Let } M \trianglelefteq G. \text{ Then, } G/M \text{ is simple iff } M \text{ is a maximal normal subgroup.} \nonumber\]

Proof

Let $\pi:G \rightarrow G/M \ (g \mapsto gM)$ and $G/M$ be simple, i.e., there are only two normal subgroups $\{e\}$ and $G/M$. By one-to-one correspondence, there are only two normal subgroup of $G$ containing $M$: $M$ and $G$. This implies that $M$ is a maximal normal subgroup of $G$.