Difference between revisions of "Aufgaben:Problem 3"
From Ferienserie MMP2
m (put the neutral element before the inverse element) |
|||
| (12 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
Let \(G\) be a finite group. For a given \(g \in G\) we consider the map \(L_g : G \rightarrow G, g' \mapsto gg'\). | 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 | + | 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 \(G\). | all invertible maps from \(G\) to \(G\). | ||
b) Prove that the map \(L\) is injective. | b) Prove that the map \(L\) is injective. | ||
| − | c) Prove that composing maps in \( | + | 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. | d) Prove that the map \(L\) is a homomorphism of groups. | ||
| − | e) Conclude that every finite group \(G\) can be considered a subgroup of \( | + | e) Conclude that every finite group \(G\) can be considered a subgroup of \(\mathrm{Sym}G\). |
| + | ==Solution== | ||
| + | [[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 15:44, 8 June 2015 (CEST) | ||
---- | ---- | ||
| − | |||
a) define \(L_g^{-1} := L_{g^{-1}}\). Now | 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) | |
| − | # | + | |
| − | # existance of neutral element: Let \(Id\) be the identity map from \(G\) to \(G\). \(\forall g \in G\) and \(\forall R \in | + | c) <span style="color: red">The numbers don't accord any more with the ones in the discussion!</span> |
| − | # existance of inverse element: by definition | + | |
| − | # | + | let \(R,S,T\) be arbitrary elements of \(SymG\) |
| + | |||
| + | # \(\forall g \in G, h:=S(g) \in G \;\Rightarrow R \circ S(g) = R(h) \in G \Rightarrow (R \circ S): G \rightarrow G\) is well defined | ||
| + | # associativity: (The composition of functions is gerenerally associative) \(\forall g \in G\) and \(\forall R, S, T \in \mathrm{Sym}G\): | ||
| + | #: $$((R \circ S) \circ T) (g) = (R \circ S) (T(g)) = R(S(T(g)))$$ | ||
| + | #: $$(R \circ (S \circ T)) (g) = R ((S \circ T)(g)) = R(S(T(g)))$$ | ||
| + | # existance of neutral element: Let \(Id\) be the identity map from \(G\) to \(G\). \(\forall g \in G\) and \(\forall R \in \mathrm{Sym}G\): \(h:=R(g), Id \circ R(g) = Id(h) = h = R (g) = R \circ Id(g)\) | ||
| + | # existance of inverse element: by definition as the identity map is the neutral element of \(SymG\) | ||
| + | # 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) | d) | ||
| − | # \(L\) is | + | # \(L\) is homomorphous. Proof: \( \forall g,h \in G: L_{g*h}(t)=g*h*t=g*L_h(t)=L_g \circ L_h (t)\) |
| − | # \(\{L_g: g \in G\}\) is a group under composition. Proof: \(\{L_g: g \in G\} \subset | + | # 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 \(\{L_g: g \in G\}\) | |
Latest revision as of 08:29, 5 August 2015
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 \(SymG\)
- \(\forall g \in G, h:=S(g) \in G \;\Rightarrow R \circ S(g) = R(h) \in G \Rightarrow (R \circ S): G \rightarrow G\) is well defined
- associativity: (The composition of functions is gerenerally associative) \(\forall g \in G\) and \(\forall R, S, T \in \mathrm{Sym}G\):
- $$((R \circ S) \circ T) (g) = (R \circ S) (T(g)) = R(S(T(g)))$$
- $$(R \circ (S \circ T)) (g) = R ((S \circ T)(g)) = R(S(T(g)))$$
- existance of neutral element: Let \(Id\) be the identity map from \(G\) to \(G\). \(\forall g \in G\) and \(\forall R \in \mathrm{Sym}G\): \(h:=R(g), Id \circ R(g) = Id(h) = h = R (g) = R \circ Id(g)\)
- existance of inverse element: by definition as the identity map is the neutral element of \(SymG\)
- 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)
- \(L\) is homomorphous. Proof: \( \forall g,h \in G: L_{g*h}(t)=g*h*t=g*L_h(t)=L_g \circ L_h (t)\)
- 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 \(\{L_g: g \in G\}\)
