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[parent] group actions and homomorphisms (Derivation)

Notes on group actions and homomorphisms

Let $ G$ be a group, $ X$ a non-empty set and $ S_X$ the symmetric group of $ X$, i.e. the group of all bijective maps on $ X$. $ \cdot$ may denote a left group action of $ G$ on $ X$.

  1. For each $ g \in G$ and $ x \in X$ we define
    $\displaystyle f_g\colon X \to X, x \mapsto g\cdot x$   .
    Since $ f_{g^-1}(f_g(x)) =g^{-1} \cdot (g \cdot x) =x$ for each $ x \in X$, $ f_{g^-1}$ is the inverse of $ f_g$. so $ f_g$ is bijective and thus element of $ S_X$. We define $ F: G \to S_X, F(g) =f_g$ for all $ g \in G$. This mapping is a group homomorphism: Let $ g,h \in G, x \in X$. Then
    $\displaystyle F(gh)(x)$ $\displaystyle =f_{gh}(x) =(gh) \cdot x =g \cdot (h \cdot x)$    
      $\displaystyle =(f_g \circ f_h)(x) =(F(g) \circ F(h))(x)$    

    for all $ x\in X$ implies $ F(gh) =F(g) \circ F(h)$. -- The same is obviously true for a right group action.
  2. Now let $ F: G \to S_x$ be a group homomorphism, and let $ f: G \times X \to X, (g,x) \mapsto F(g)(x)$ satisfy
    1. $ f(1_G, x) =F(1_g)(x) =x$ for all $ x \in X$ and
    2. $ f(gh, x) =F(gh)(x) =(F(g) \circ F(h)(x) =F(g)(F(h)(x)) =f(g, f(h,x))$,
    so $ f$ is a group action induced by $ F$.

Characterization of group actions

Let $ G$ be a group acting on a set $ X$. Using the same notation as above, we have for each $ g \in \operatorname{ker}(F)$
$\displaystyle F(g) =\operatorname{id}_x =f_g \Leftrightarrow g \cdot x =x \quad \forall \quad x \in X \Leftrightarrow g \in \cup_{x \in X} G_x$ (1)

and it follows
$\displaystyle \operatorname{ker}(F) =\bigcap_{x \in X} G_x. $
Let $ G$ act transitively on $ X$. Then for any $ x \in X$ $ X$ is the orbit $ G(x)$ of $ x$. As shown in “conjugate stabilizer subgroups', all stabilizer subgroups of elements $ y \in G(x)$ are conjugate subgroups to $ G_x$ in $ G$. From the above it follows that
$\displaystyle \operatorname{ker}(F) =\bigcap_{g \in G} gG_xg^{-1}. $
For a faithful operation of $ G$ the condition $ g \cdot x =x \quad \forall \quad x \in X \rightarrow g =1_G$ is equivalent to
$\displaystyle \operatorname{ker}(F) =\{1_G\} $
and therefore $ F\colon G \to S_X$ is a monomorphism.

For the trivial operation of $ G$ on $ X$ given by $ g \cdot x =x \forall g \in G$ the stabilizer subgroup $ G_x$ is $ G$ for all $ x \in X$, and thus

$\displaystyle \operatorname{ker}(F) =G. $

If the operation of $ G$ on $ X$ is free, then $ G_x =\{1_G\} \;\forall \; x \in X$, thus the kernel of $ F$ is $ \{1_G\}$-like for a faithful operation. But:

Let $ X=\{1, \ldots, n\}$ and $ G=S_n$. Then the operation of $ G$ on $ X$ given by

$\displaystyle \pi \cdot i := \pi(i) \forall i \in X, \pi \in S_n $
is faithful but not free.



"group actions and homomorphisms" is owned by CWoo. [ full author list (2) | owner history (6) ]
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See Also: group homomorphism

Keywords:  SymmetricGroup, GroupHomomorphism

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Cross-references: kernel, monomorphism, equivalent, operation, faithful, conjugate subgroups, subgroups, stabilizer, conjugate stabilizer subgroups, orbit, induced, right, implies, group homomorphism, mapping, inverse, group action, maps, bijective, symmetric group, group
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This is version 11 of group actions and homomorphisms, born on 2002-12-23, modified 2005-03-14.
Object id is 3820, canonical name is GroupActionsAndHomomorphisms.
Accessed 3305 times total.

Classification:
AMS MSC: 20A05 (Group theory and generalizations :: Foundations :: Axiomatics and elementary properties)
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