Prove subset of transformations is a group
0
I'm trying to prove that the subset of linear transformations given below is a group:
Let $T^n$ denote the n-fold composition of any $T∈ℒ(V)$. For instance, the two fold composition $Tcirc T$ is denoted $T^2$.
Let $T∈ℒ(V)$ such that $T≠I$ and $T^n=I$. Show that $C_n={I^{},T^{},T^2,...T^{n−1}}$ is a group.
linear-algebra abstract-algebra group-theory cyclic-groups
asked Dec 2 '18 at 23:21
courtorder52
11
|
show 4 more comments
0
I'm trying to prove that the subset of linear transformations given below is a group:
Let $T^n$ denote the n-fold composition of any $T∈ℒ(V)$. For instance, the two fold composition $Tcirc T$ is denoted $T^2$.
Let $T∈ℒ(V)$ such that $T≠I$ and $T^n=I$. Show that $C_n={I^{},T^{},T^2,...T^{n−1}}$ is a group.
linear-algebra abstract-algebra group-theory cyclic-groups
asked Dec 2 '18 at 23:21
courtorder52
11
Which of the group axioms can't you prove?
– Rob Arthan
Dec 2 '18 at 23:26
Closure under composition
– courtorder52
Dec 2 '18 at 23:35
Given $T^n=I$, that shouldn't be too difficult though. Note that $T^iT^j=T^{i+j}$, because of associativity of composition.
– Jean-Claude Arbaut
Dec 2 '18 at 23:36
Why though? I'm taking $T^n$ to refer to a specific transformation, not as the definition for all transformations.
– courtorder52
Dec 2 '18 at 23:40
Sure. You are wondering why the set is closed under composition. Prove that the composition of two of its elements is still in it. But the elements can be written $T^i$ and $T^j$, and the composition is $T^{i+j}$. Either $i+jle n$ and there is nothing to prove, either $i+j>n$ but then $T^{i+j}=T^nT^{i+j-n}$. Is it clearer now?
– Jean-Claude Arbaut
Dec 2 '18 at 23:45
|
show 4 more comments
0
0
0
I'm trying to prove that the subset of linear transformations given below is a group:
Let $T^n$ denote the n-fold composition of any $T∈ℒ(V)$. For instance, the two fold composition $Tcirc T$ is denoted $T^2$.
Let $T∈ℒ(V)$ such that $T≠I$ and $T^n=I$. Show that $C_n={I^{},T^{},T^2,...T^{n−1}}$ is a group.
linear-algebra abstract-algebra group-theory cyclic-groups
asked Dec 2 '18 at 23:21
courtorder52
11
I'm trying to prove that the subset of linear transformations given below is a group:
Let $T^n$ denote the n-fold composition of any $T∈ℒ(V)$. For instance, the two fold composition $Tcirc T$ is denoted $T^2$.
Let $T∈ℒ(V)$ such that $T≠I$ and $T^n=I$. Show that $C_n={I^{},T^{},T^2,...T^{n−1}}$ is a group.
linear-algebra abstract-algebra group-theory cyclic-groups
linear-algebra abstract-algebra group-theory cyclic-groups
asked Dec 2 '18 at 23:21
courtorder52
11
asked Dec 2 '18 at 23:21
courtorder52
11
asked Dec 2 '18 at 23:21
courtorder52
11
asked Dec 2 '18 at 23:21
courtorder52
11
asked Dec 2 '18 at 23:21
courtorder52
11
11
Which of the group axioms can't you prove?
– Rob Arthan
Dec 2 '18 at 23:26
Closure under composition
– courtorder52
Dec 2 '18 at 23:35
Given $T^n=I$, that shouldn't be too difficult though. Note that $T^iT^j=T^{i+j}$, because of associativity of composition.
– Jean-Claude Arbaut
Dec 2 '18 at 23:36
Why though? I'm taking $T^n$ to refer to a specific transformation, not as the definition for all transformations.
– courtorder52
Dec 2 '18 at 23:40
Sure. You are wondering why the set is closed under composition. Prove that the composition of two of its elements is still in it. But the elements can be written $T^i$ and $T^j$, and the composition is $T^{i+j}$. Either $i+jle n$ and there is nothing to prove, either $i+j>n$ but then $T^{i+j}=T^nT^{i+j-n}$. Is it clearer now?
– Jean-Claude Arbaut
Dec 2 '18 at 23:45
|
show 4 more comments
Which of the group axioms can't you prove?
– Rob Arthan
Dec 2 '18 at 23:26
Closure under composition
– courtorder52
Dec 2 '18 at 23:35
Given $T^n=I$, that shouldn't be too difficult though. Note that $T^iT^j=T^{i+j}$, because of associativity of composition.
– Jean-Claude Arbaut
Dec 2 '18 at 23:36
Why though? I'm taking $T^n$ to refer to a specific transformation, not as the definition for all transformations.
– courtorder52
Dec 2 '18 at 23:40
Sure. You are wondering why the set is closed under composition. Prove that the composition of two of its elements is still in it. But the elements can be written $T^i$ and $T^j$, and the composition is $T^{i+j}$. Either $i+jle n$ and there is nothing to prove, either $i+j>n$ but then $T^{i+j}=T^nT^{i+j-n}$. Is it clearer now?
– Jean-Claude Arbaut
Dec 2 '18 at 23:45
Which of the group axioms can't you prove?
– Rob Arthan
Dec 2 '18 at 23:26
Which of the group axioms can't you prove?
– Rob Arthan
Dec 2 '18 at 23:26
Closure under composition
– courtorder52
Dec 2 '18 at 23:35
Closure under composition
– courtorder52
Dec 2 '18 at 23:35
Given $T^n=I$, that shouldn't be too difficult though. Note that $T^iT^j=T^{i+j}$, because of associativity of composition.
– Jean-Claude Arbaut
Dec 2 '18 at 23:36
Given $T^n=I$, that shouldn't be too difficult though. Note that $T^iT^j=T^{i+j}$, because of associativity of composition.
– Jean-Claude Arbaut
Dec 2 '18 at 23:36
Why though? I'm taking $T^n$ to refer to a specific transformation, not as the definition for all transformations.
– courtorder52
Dec 2 '18 at 23:40
Why though? I'm taking $T^n$ to refer to a specific transformation, not as the definition for all transformations.
– courtorder52
Dec 2 '18 at 23:40
Sure. You are wondering why the set is closed under composition. Prove that the composition of two of its elements is still in it. But the elements can be written $T^i$ and $T^j$, and the composition is $T^{i+j}$. Either $i+jle n$ and there is nothing to prove, either $i+j>n$ but then $T^{i+j}=T^nT^{i+j-n}$. Is it clearer now?
– Jean-Claude Arbaut
Dec 2 '18 at 23:45
Sure. You are wondering why the set is closed under composition. Prove that the composition of two of its elements is still in it. But the elements can be written $T^i$ and $T^j$, and the composition is $T^{i+j}$. Either $i+jle n$ and there is nothing to prove, either $i+j>n$ but then $T^{i+j}=T^nT^{i+j-n}$. Is it clearer now?
– Jean-Claude Arbaut
Dec 2 '18 at 23:45
|
show 4 more comments
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Which of the group axioms can't you prove?
– Rob Arthan
Dec 2 '18 at 23:26
Closure under composition
– courtorder52
Dec 2 '18 at 23:35
Given $T^n=I$, that shouldn't be too difficult though. Note that $T^iT^j=T^{i+j}$, because of associativity of composition.
– Jean-Claude Arbaut
Dec 2 '18 at 23:36
Why though? I'm taking $T^n$ to refer to a specific transformation, not as the definition for all transformations.
– courtorder52
Dec 2 '18 at 23:40
Sure. You are wondering why the set is closed under composition. Prove that the composition of two of its elements is still in it. But the elements can be written $T^i$ and $T^j$, and the composition is $T^{i+j}$. Either $i+jle n$ and there is nothing to prove, either $i+j>n$ but then $T^{i+j}=T^nT^{i+j-n}$. Is it clearer now?
– Jean-Claude Arbaut
Dec 2 '18 at 23:45