Aufgaben:Problem 3

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Problem

Let \(G\) be a finite group. For a given \(g \in G\) we consider the map \(L_g : G \rightarrow G, g' \mapsto gg'\).

a) Prove that \(L : g \mapsto L_g\) defines a map \(G \rightarrow \mathrm{Sym}G\) where \(\mathrm{Sym}G\) denotes the set of all invertible maps from \(G\) to \(G\).

b) Prove that the map \(L\) is injective.

c) Prove that composing maps in \(\mathrm{Sym}G\) defines a group structure on \(\mathrm{Sym}G\).

d) Prove that the map \(L\) is a homomorphism of groups.

e) Conclude that every finite group \(G\) can be considered a subgroup of \(\mathrm{Sym}G\).

Solution

Brynerm (talk) 15:44, 8 June 2015 (CEST)


a) define \(L_g^{-1} := L_{g^{-1}}\). Now

$$\forall h \in G: L_g^{-1} \circ L_g(h)=L_{g^{-1}}(g*h)=g^{-1}gh=h \;\Rightarrow L_{g^{-1}} \circ L_g=Id$$ $$\forall h \in G: L_g \circ L_g^{-1}(h)=L_g(g^{-1}*h)=gg^{-1}h=h \;\Rightarrow L_g \circ L_{g^{-1}}=Id$$

That holds that all \(L_g\) are invertible.

b) Assume \(L(g)=L(h) \Rightarrow L_g(t)=L_h(t), \forall t \in G\;\Rightarrow gt=ht, \forall t \;\Rightarrow g=h\) (as \(t\in G\) is invertible)

c) The numbers don't accord any more with the ones in the discussion!

let \(R,S,T\) be arbitrary elements of \(SymG\)

  1. \(\forall g \in G, h:=S(g) \in G \;\Rightarrow R \circ S(g) = R(h) \in G \Rightarrow (R \circ S): G \rightarrow G\) is well defined
  2. associativity: (The composition of functions is gerenerally associative) \(\forall g \in G\) and \(\forall R, S, T \in \mathrm{Sym}G\):
    $$((R \circ S) \circ T) (g) = (R \circ S) (T(g)) = R(S(T(g)))$$
    $$(R \circ (S \circ T)) (g) = R ((S \circ T)(g)) = R(S(T(g)))$$
  3. existance of neutral element: Let \(Id\) be the identity map from \(G\) to \(G\). \(\forall g \in G\) and \(\forall R \in \mathrm{Sym}G\): \(h:=R(g), Id \circ R(g) = Id(h) = h = R (g) = R \circ Id(g)\)
  4. existance of inverse element: by definition as the identity map is the neutral element of \(SymG\)
  5. closure: \(\forall R, S\in \mathrm{Sym}G\): \((R \circ S) \in \mathrm{Sym}G\). Proof:
    \((S^{-1} \circ R^{-1}) \in \mathrm{Sym}G\) is the inverse element of \((R \circ S) \), because
    $$\forall g \in G: ((S^{-1} \circ R^{-1}) \circ (R \circ S))(g)=(S^{-1} \circ R^{-1} \circ R \circ S)(g)=(S^{-1} \circ S)(g)=Id(g)$$
    $$\forall g \in G: ((R \circ S) \circ (S^{-1} \circ R^{-1}))(g)=(R \circ S \circ S^{-1} \circ R^{-1})(g)=(R \circ R^{-1})(g)=Id(g)$$


d)

  1. \(L\) is homomorphous. Proof: \( \forall g,h \in G: L_{g*h}(t)=g*h*t=g*L_h(t)=L_g \circ L_h (t)\)
  2. from c) we know that \(symG\) has group structure, therefore \(L\) is homomorphism of groups


e)

The map \(L\) is defined for any finite group.

\(\{L_g: g \in G\}\) is a group under composition. Proof:

  • \(\{L_g: g \in G\}\neq\{\}\) as \(G\neq\{\}\)
  • \(\{L_g: g \in G\}\subset \mathrm{Sym}G\) and \(L_g^{-1} \in \{L_g: g \in G\}\) as shown in a)
  • \(\forall L_g, L_h \in \{L_g: g \in G\}: (L_g \circ L_h) \in \{L_g: g \in G\}\) Proof: from 1.) \(\Rightarrow \forall g,h \in G: L_g \circ L_h = L_{g*h} = L_{t}\) , with \(t=g*h \in G \)
\(\Rightarrow (\{L_g: g \in G\},\circ)\) is a subgroup of \((\mathrm{Sym}G ,\circ)\)

\(L\) is injective and homomorphous and if we restrict \(L\) to its image it becomes also surjective. So \(L^*: G \rightarrow L(G)=\{L_g: g \in G\} \subset \mathrm{Sym}G\) is a group isomorphism. Therefore every group \(G\) is isomorphic to \(\{L_g: g \in G\}\)