Difference between revisions of "Aufgaben:Problem 3"
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# \(\{L_g: g \in G\}\) is a group under composition. Proof: \(\{L_g: g \in G\} \subset SymG\) as shown above and from 1.) fallows \(L_{g*h} = L_{t}\) , with \(t=g*h \in G \Rightarrow L_{g*h} \in \{L_g: g \in G\} \Rightarrow \{L_g: g \in G\}\) is a subgroup of \(SymG\) with composition | # \(\{L_g: g \in G\}\) is a group under composition. Proof: \(\{L_g: g \in G\} \subset SymG\) as shown above and from 1.) fallows \(L_{g*h} = L_{t}\) , with \(t=g*h \in G \Rightarrow L_{g*h} \in \{L_g: g \in G\} \Rightarrow \{L_g: g \in G\}\) is a subgroup of \(SymG\) with composition | ||
| − | e) The map \(L | + | e) The map \(L\) is defined for any finite group. As \(L\) is injective and homomorph, \(L\) is a group isomorphism. Therefore every group \(G\) is isomorph to \(L(G)=\{L_g: g \in G\}\) |
Revision as of 14:04, 8 June 2015
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 \(SymG\).
--Brynerm (talk) 15:44, 8 June 2015 (CEST)
a) define \(L_g^{-1} := L_{g^{-1}}\). Now \( \forall h: L_g^{-1} \circ L_g(h)=L_{g^{-1}}(g*h)=g^{-1}gh=h \;\Rightarrow L_{g^{-1}} \circ L_g=Id \) That holds that all \(L_g\) are invertible
b) Assume \(L(g)=L(h) \Rightarrow L_g(t)=L_h(t), \forall t \;\Rightarrow gt=ht, \forall t \;\Rightarrow g=h\) (as \(t\in G\) is invertible)
c)
- \(\forall g \in G\) and \(R, S\in SymG\): \(h:=S(g) \in G \;\Rightarrow R \circ S(g) = R(h) \in G \)
- existance of neutral element: Let \(Id\) be the identity map from \(G\) to \(G\). \(\forall g \in G\) and \(\forall R \in SymG\): \(h:=R(g), Id \circ R(g) = Id(h) = h = R (g) = R \circ Id(g)\)
- existance of inverse element: by definition
- associativity: \(\forall g \in G\) 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)
- \(L\) is homomoph. Proof: \(L_{g*h}(t)=g*h*t=g*L_h(t)=L_g \circ L_h (t)\)
- \(\{L_g: g \in G\}\) is a group under composition. Proof: \(\{L_g: g \in G\} \subset SymG\) as shown above and from 1.) fallows \(L_{g*h} = L_{t}\) , with \(t=g*h \in G \Rightarrow L_{g*h} \in \{L_g: g \in G\} \Rightarrow \{L_g: g \in G\}\) is a subgroup of \(SymG\) with composition
e) The map \(L\) is defined for any finite group. As \(L\) is injective and homomorph, \(L\) is a group isomorphism. Therefore every group \(G\) is isomorph to \(L(G)=\{L_g: g \in G\}\)
