Aufgaben:Problem 3

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 \(G \rightarrow SymG\) where \(SymG\) 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 \(SymG\) defines a group structure on \(SymG\).

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

e) Conclude that every finite group \(G\) can be considered a subgroup of \(SymG\).

Solution

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

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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)

1. \(\forall R, S\in SymG\): \((R \circ S) \in SymG\). Proof:
   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. \((S^{-1} \circ R^{-1}) \in SymG\) is the inverse element of \((R \circ S) \), because

      $$\forall g \in G: ((S^{-1} \circ R^{-1}) \circ (R \circ S))(g)=(S^{-1} \circ R^{-1} \circ R \circ S)(g)=(S^{-1} \circ S)(g)=Id(g)$$

      $$\forall g \in G: ((R \circ S) \circ (S^{-1} \circ R^{-1}))(g)=(R \circ S \circ S^{-1} \circ R^{-1})(g)=(R \circ R^{-1})(g)=Id(g)$$

2. existance of neutral element: Let \(Id\) be the identity map from \(G\) to \(G\). \(\forall g \in G\) and \(\forall R \in SymG\): \(h:=R(g), Id \circ R(g) = Id(h) = h = R (g) = R \circ Id(g)\)
3. existance of inverse element: by definition
4. associativity: (The composition of functions is gerenerally associative) \(\forall g \in G\) and \(\forall R, S, T \in SymG\):

   $$((R \circ S) \circ T) (g) = (R \circ S) (T(g)) = R(S(T(g)))$$

   $$(R \circ (S \circ T)) (g) = R ((S \circ T)(g)) = R(S(T(g)))$$

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. \(\{L_g: g \in G\}\) is a group under composition. Proof:
   • \(\{L_g: g \in G\}\neq\{\}\) as \(G\neq\{\}\)
   • \(\{L_g: g \in G\}\subset SymG\) 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 \((SymG ,\circ)\)

e) The map \(L\) is defined for any finite group. As \(L\) is injective and homomorphous, \(L: G \rightarrow \{L_g: g \in G\} \subset SymG\) is a group isomorphism. Therefore every group \(G\) is isomorphic to \(\{L_g: g \in G\}\)