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Ali Taghavi
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Let $G$ be a topological group. We define an equivalence relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the following two maps are topologically conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

Obviously the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalence classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalence relation is the following: two elements are equivalent if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalence class containing $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalence class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalence relations on finite groups enable us to obtain some useful counting theorems. Even Sylow's theorems were direct or undirect consequence of some equivalence relations. So I wonder is this equivalence relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalence classes. On the other hand for circle case every equivalence class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalence class approach to zero or approach to 1 (w.r.t. the Haar measure)? Is every equivalence class necessarily measurable? the 2 last questionThat means: can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example.

However one should first answer to: Is every equivalence class necessarily measurable?

Let $G$ be a topological group. We define an equivalence relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the following two maps are topologically conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

Obviously the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalence classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalence relation is the following: two elements are equivalent if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalence class containing $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalence class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalence relations on finite groups enable us to obtain some useful counting theorems. Even Sylow's theorems were direct or undirect consequence of some equivalence relations. So I wonder is this equivalence relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalence classes. On the other hand for circle case every equivalence class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalence class approach to zero or approach to 1 (w.r.t. the Haar measure)? Is every equivalence class necessarily measurable? the 2 last question means: can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example.

Let $G$ be a topological group. We define an equivalence relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the following two maps are topologically conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

Obviously the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalence classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalence relation is the following: two elements are equivalent if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalence class containing $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalence class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalence relations on finite groups enable us to obtain some useful counting theorems. Even Sylow's theorems were direct or undirect consequence of some equivalence relations. So I wonder is this equivalence relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalence classes. On the other hand for circle case every equivalence class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalence class approach to zero or approach to 1 (w.r.t. the Haar measure)? That means: can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example.

However one should first answer to: Is every equivalence class necessarily measurable?

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YCor
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Let $G$ be a topologicala topological group. We define anan equivalence relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the followingfollowing two maps are topologicaltopologically conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

ObviouselyObviously the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalence classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalence relation is the following: two elements are equivalentsequivalent if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalence class containigcontaining $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalence class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalence relations on finite groups enable us to obtain some useful counting theorems. Even SylowsSylow's theorems were direct or undirect consequence of some equivalence relations. So I wonder is this equivalence relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalence classes. On the other hand for circle case every equivalence class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalence class approach to zero or approach to 1 (wrtw.r.t. the Haar measure)?  Is every equivalence class necessarillynecessarily measurable? the 2 last question means: can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example.

Let $G$ be a topological group. We define an equivalence relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the following two maps are topological conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

Obviousely the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalence classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalence relation is the following: two elements are equivalents if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalence class containig $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalence class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalence relations on finite groups enable us to obtain some useful counting theorems. Even Sylows theorems were direct or undirect consequence of some equivalence relations. So I wonder is this equivalence relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalence classes. On the other hand for circle case every equivalence class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalence class approach to zero or approach to 1 (wrt the Haar measure)?Is every equivalence class necessarilly measurable? the 2 last question means: can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example

Let $G$ be a topological group. We define an equivalence relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the following two maps are topologically conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

Obviously the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalence classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalence relation is the following: two elements are equivalent if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalence class containing $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalence class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalence relations on finite groups enable us to obtain some useful counting theorems. Even Sylow's theorems were direct or undirect consequence of some equivalence relations. So I wonder is this equivalence relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalence classes. On the other hand for circle case every equivalence class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalence class approach to zero or approach to 1 (w.r.t. the Haar measure)?  Is every equivalence class necessarily measurable? the 2 last question means: can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example.

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Ali Taghavi
  • 356
  • 8
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  • 123

Let $G$ be a topological group. We define an equivalent relation equivalence relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the following two maps are topological conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

Obviousely the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalent equivalence classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalentequivalence relation is the following: two elements are equivalents if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalentequivalence class containig $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalent equivalence class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalents equivalence relations on finite groups enable us to obtain some useful counting theorems. Even Sylows theoremtheorems were direct or undirect consequence of some equivalentequivalence relations. So I wonder is this equivalent equivalence relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalent classequivalence classes. On the other hand for circle case every equivalentequivalence class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalentequivalence class approach to zero or approach to 1 (thewrt the Haar measure)? Is every equivalent class necessarilly measurable?Is every equivalence class necessarilly measurable? I meanthe 2 last question means: can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example

Let $G$ be a topological group. We define an equivalent relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the following two maps are topological conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

Obviousely the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalent classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalent relation is the following: two elements are equivalents if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalent class containig $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalent class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalents relations on finite groups enable us to obtain useful counting theorems. Even Sylows theorem were direct or undirect consequence of some equivalent relations. So I wonder is this equivalent relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalent class. On the other hand for circle case every equivalent class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalent class approach to zero or approach to 1 (the Haar measure)? Is every equivalent class necessarilly measurable? I mean can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example

Let $G$ be a topological group. We define an equivalence relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the following two maps are topological conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

Obviousely the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalence classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalence relation is the following: two elements are equivalents if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalence class containig $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalence class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalence relations on finite groups enable us to obtain some useful counting theorems. Even Sylows theorems were direct or undirect consequence of some equivalence relations. So I wonder is this equivalence relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalence classes. On the other hand for circle case every equivalence class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalence class approach to zero or approach to 1 (wrt the Haar measure)?Is every equivalence class necessarilly measurable? the 2 last question means: can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example

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Ali Taghavi
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Ali Taghavi
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