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Real valued function concatenation as group operator? For which sets?
Let $G,*$ and $G',circ$ be groups, prove that $S$ forms a group under the product operationProving that something is a groupShow that $G$ is a groupAssociativity of function composition?Proving that [R, +] forms a groupDetermining whether a binary operation is a group.Does having exactly one of each element in every row and column of a Cayley table ensure that it’s a groupWhy is group theory axiomatized with operations?Prove or disprove that the set of continuous mappings $phicolon [0;1]to [0;1]$ such that […] forms a group.Does a set of all real valued functions form group under componentwise multiplication?
$begingroup$
Can we define function concatenation of normal functions $$xin mathbb R\xto f(x)inmathbb R$$ as group operator ?
The identity element would be $f(x) = x$, I suppose
Function inverse can be defined as $f^-1(x)$ s.t. $f^-1(f(x)) = 0$.
In group setting I suppose our group's inverse element could simply be this function inverse?
These are the two seemingly uncontested properties which fit well.
But... what about associativity? $(f circ g) circ h = f circ (g circ h)$. Are we sure this will be satisfied?
And what about closure? What families of functions would guarantee this? One example should be polynomials, but I am quite sure that inverse element would often not exist within polynomials...
real-analysis abstract-algebra group-theory functional-analysis soft-question
asked Mar 28 at 15:08
mathreadlermathreadler
15.4k72263
$endgroup$
|
show 5 more comments
$begingroup$
Can we define function concatenation of normal functions $$xin mathbb R\xto f(x)inmathbb R$$ as group operator ?
The identity element would be $f(x) = x$, I suppose
Function inverse can be defined as $f^-1(x)$ s.t. $f^-1(f(x)) = 0$.
In group setting I suppose our group's inverse element could simply be this function inverse?
These are the two seemingly uncontested properties which fit well.
But... what about associativity? $(f circ g) circ h = f circ (g circ h)$. Are we sure this will be satisfied?
And what about closure? What families of functions would guarantee this? One example should be polynomials, but I am quite sure that inverse element would often not exist within polynomials...
real-analysis abstract-algebra group-theory functional-analysis soft-question
asked Mar 28 at 15:08
mathreadlermathreadler
15.4k72263
$endgroup$
3
$begingroup$
The functions would obviously need to be bijective, but other than that, this should hold.
$endgroup$
– Don Thousand
Mar 28 at 15:10
1
$begingroup$
I don't think polynomials are a good place to look since many are non-invertible (as many are not bijective).
$endgroup$
– Don Thousand
Mar 28 at 15:17
1
$begingroup$
Well, there's the group of order two consisting of the functions $x mapsto x$ and $x mapsto 1 - x$. If that's not exciting enough, and if one is allowed to compactify $mathbbR$ into $overlinemathbbR = mathbbR cup infty$, then one can enlarge that example into the real version of the anharmonic group, consisting of six functions $x mapsto x$, $x mapsto 1 - x$, $x mapsto frac1x$, $x mapsto frac11 - x$, $x mapsto 1 - frac1x$, and $x mapsto fracxx - 1$. I'll get me coat ...
$endgroup$
– Calum Gilhooley
Mar 28 at 20:13
1
$begingroup$
Perhaps more interestingly, and not requiring $infty$, there is the group of all non-constant affine functions, viz. $x mapsto ax + b$, where $a, b in mathbbR$ and $a ne 0$. And this has another subgroup, infinite this time, consisting of just the functions $x mapsto ax$, where $a ne 0$.
$endgroup$
– Calum Gilhooley
Mar 28 at 20:29
1
$begingroup$
But if $m(x)= cx^n$, then $m^-1(x) = sqrt[n]fracxc$, and not, as you write, $1/ccdot x^-n$.
$endgroup$
– Calum Gilhooley
Mar 28 at 21:49
|
show 5 more comments
$begingroup$
Can we define function concatenation of normal functions $$xin mathbb R\xto f(x)inmathbb R$$ as group operator ?
The identity element would be $f(x) = x$, I suppose
Function inverse can be defined as $f^-1(x)$ s.t. $f^-1(f(x)) = 0$.
In group setting I suppose our group's inverse element could simply be this function inverse?
These are the two seemingly uncontested properties which fit well.
But... what about associativity? $(f circ g) circ h = f circ (g circ h)$. Are we sure this will be satisfied?
And what about closure? What families of functions would guarantee this? One example should be polynomials, but I am quite sure that inverse element would often not exist within polynomials...
real-analysis abstract-algebra group-theory functional-analysis soft-question
asked Mar 28 at 15:08
mathreadlermathreadler
15.4k72263
$endgroup$
Can we define function concatenation of normal functions $$xin mathbb R\xto f(x)inmathbb R$$ as group operator ?
The identity element would be $f(x) = x$, I suppose
Function inverse can be defined as $f^-1(x)$ s.t. $f^-1(f(x)) = 0$.
In group setting I suppose our group's inverse element could simply be this function inverse?
These are the two seemingly uncontested properties which fit well.
But... what about associativity? $(f circ g) circ h = f circ (g circ h)$. Are we sure this will be satisfied?
And what about closure? What families of functions would guarantee this? One example should be polynomials, but I am quite sure that inverse element would often not exist within polynomials...
real-analysis abstract-algebra group-theory functional-analysis soft-question
real-analysis abstract-algebra group-theory functional-analysis soft-question
asked Mar 28 at 15:08
mathreadlermathreadler
15.4k72263
asked Mar 28 at 15:08
mathreadlermathreadler
15.4k72263
asked Mar 28 at 15:08
mathreadlermathreadler
15.4k72263
asked Mar 28 at 15:08
mathreadlermathreadler
15.4k72263
asked Mar 28 at 15:08
mathreadlermathreadler
15.4k72263
15.4k72263
3
$begingroup$
The functions would obviously need to be bijective, but other than that, this should hold.
$endgroup$
– Don Thousand
Mar 28 at 15:10
1
$begingroup$
I don't think polynomials are a good place to look since many are non-invertible (as many are not bijective).
$endgroup$
– Don Thousand
Mar 28 at 15:17
1
$begingroup$
Well, there's the group of order two consisting of the functions $x mapsto x$ and $x mapsto 1 - x$. If that's not exciting enough, and if one is allowed to compactify $mathbbR$ into $overlinemathbbR = mathbbR cup infty$, then one can enlarge that example into the real version of the anharmonic group, consisting of six functions $x mapsto x$, $x mapsto 1 - x$, $x mapsto frac1x$, $x mapsto frac11 - x$, $x mapsto 1 - frac1x$, and $x mapsto fracxx - 1$. I'll get me coat ...
$endgroup$
– Calum Gilhooley
Mar 28 at 20:13
1
$begingroup$
Perhaps more interestingly, and not requiring $infty$, there is the group of all non-constant affine functions, viz. $x mapsto ax + b$, where $a, b in mathbbR$ and $a ne 0$. And this has another subgroup, infinite this time, consisting of just the functions $x mapsto ax$, where $a ne 0$.
$endgroup$
– Calum Gilhooley
Mar 28 at 20:29
1
$begingroup$
But if $m(x)= cx^n$, then $m^-1(x) = sqrt[n]fracxc$, and not, as you write, $1/ccdot x^-n$.
$endgroup$
– Calum Gilhooley
Mar 28 at 21:49
|
show 5 more comments
3
$begingroup$
The functions would obviously need to be bijective, but other than that, this should hold.
$endgroup$
– Don Thousand
Mar 28 at 15:10
1
$begingroup$
I don't think polynomials are a good place to look since many are non-invertible (as many are not bijective).
$endgroup$
– Don Thousand
Mar 28 at 15:17
1
$begingroup$
Well, there's the group of order two consisting of the functions $x mapsto x$ and $x mapsto 1 - x$. If that's not exciting enough, and if one is allowed to compactify $mathbbR$ into $overlinemathbbR = mathbbR cup infty$, then one can enlarge that example into the real version of the anharmonic group, consisting of six functions $x mapsto x$, $x mapsto 1 - x$, $x mapsto frac1x$, $x mapsto frac11 - x$, $x mapsto 1 - frac1x$, and $x mapsto fracxx - 1$. I'll get me coat ...
$endgroup$
– Calum Gilhooley
Mar 28 at 20:13
1
$begingroup$
Perhaps more interestingly, and not requiring $infty$, there is the group of all non-constant affine functions, viz. $x mapsto ax + b$, where $a, b in mathbbR$ and $a ne 0$. And this has another subgroup, infinite this time, consisting of just the functions $x mapsto ax$, where $a ne 0$.
$endgroup$
– Calum Gilhooley
Mar 28 at 20:29
1
$begingroup$
But if $m(x)= cx^n$, then $m^-1(x) = sqrt[n]fracxc$, and not, as you write, $1/ccdot x^-n$.
$endgroup$
– Calum Gilhooley
Mar 28 at 21:49
3
3
$begingroup$
The functions would obviously need to be bijective, but other than that, this should hold.
$endgroup$
– Don Thousand
Mar 28 at 15:10
$begingroup$
The functions would obviously need to be bijective, but other than that, this should hold.
$endgroup$
– Don Thousand
Mar 28 at 15:10
1
1
$begingroup$
I don't think polynomials are a good place to look since many are non-invertible (as many are not bijective).
$endgroup$
– Don Thousand
Mar 28 at 15:17
$begingroup$
I don't think polynomials are a good place to look since many are non-invertible (as many are not bijective).
$endgroup$
– Don Thousand
Mar 28 at 15:17
1
1
$begingroup$
Well, there's the group of order two consisting of the functions $x mapsto x$ and $x mapsto 1 - x$. If that's not exciting enough, and if one is allowed to compactify $mathbbR$ into $overlinemathbbR = mathbbR cup infty$, then one can enlarge that example into the real version of the anharmonic group, consisting of six functions $x mapsto x$, $x mapsto 1 - x$, $x mapsto frac1x$, $x mapsto frac11 - x$, $x mapsto 1 - frac1x$, and $x mapsto fracxx - 1$. I'll get me coat ...
$endgroup$
– Calum Gilhooley
Mar 28 at 20:13
$begingroup$
Well, there's the group of order two consisting of the functions $x mapsto x$ and $x mapsto 1 - x$. If that's not exciting enough, and if one is allowed to compactify $mathbbR$ into $overlinemathbbR = mathbbR cup infty$, then one can enlarge that example into the real version of the anharmonic group, consisting of six functions $x mapsto x$, $x mapsto 1 - x$, $x mapsto frac1x$, $x mapsto frac11 - x$, $x mapsto 1 - frac1x$, and $x mapsto fracxx - 1$. I'll get me coat ...
$endgroup$
– Calum Gilhooley
Mar 28 at 20:13
1
1
$begingroup$
Perhaps more interestingly, and not requiring $infty$, there is the group of all non-constant affine functions, viz. $x mapsto ax + b$, where $a, b in mathbbR$ and $a ne 0$. And this has another subgroup, infinite this time, consisting of just the functions $x mapsto ax$, where $a ne 0$.
$endgroup$
– Calum Gilhooley
Mar 28 at 20:29
$begingroup$
Perhaps more interestingly, and not requiring $infty$, there is the group of all non-constant affine functions, viz. $x mapsto ax + b$, where $a, b in mathbbR$ and $a ne 0$. And this has another subgroup, infinite this time, consisting of just the functions $x mapsto ax$, where $a ne 0$.
$endgroup$
– Calum Gilhooley
Mar 28 at 20:29
1
1
$begingroup$
But if $m(x)= cx^n$, then $m^-1(x) = sqrt[n]fracxc$, and not, as you write, $1/ccdot x^-n$.
$endgroup$
– Calum Gilhooley
Mar 28 at 21:49
$begingroup$
But if $m(x)= cx^n$, then $m^-1(x) = sqrt[n]fracxc$, and not, as you write, $1/ccdot x^-n$.
$endgroup$
– Calum Gilhooley
Mar 28 at 21:49
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
You only need closure under composition and inverses as composition of functions is always associative.
You may prove this as follows: Let $f:Xto Y$, $g:Yto Z$, $h:Zto W$ be functions with $X,Y,Z,W$ non-empty sets. Then, for any $xin X$, we have
$$((hcirc g)circ f)(x)=(hcirc g)(f(x))=h(g(f(x)))=h((gcirc f)(x))=(hcirc(gcirc f))(x)$$
and so $(hcirc g)circ f$ and $hcirc(gcirc f)$ are pointwise equal and thus equal as functions.
Note that this "proof" is really just analysis the nature of composition as we just shift the brackets around; there is no hidden meaning; composition is inherently associative.
An example would the space of all continuous functions from $mathbb R$ to $mathbb R$ which are invertible, or in other words bijective. Composition of continuous functions gives you a continuous function and the composition of two bijective functions is bijective.
Another classical example is the group of automorphism of an algebraic structure, e.g. for a vector space $V$ the set of all bijective linear maps $phi: Vto V$.
answered Mar 28 at 15:15
blubblub
2,770826
$endgroup$
$begingroup$
This is what I hoped and thought, but I did not remember any proof of associativity so I was not sure.
$endgroup$
– mathreadler
Mar 28 at 15:17
add a comment |
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$begingroup$
You only need closure under composition and inverses as composition of functions is always associative.
You may prove this as follows: Let $f:Xto Y$, $g:Yto Z$, $h:Zto W$ be functions with $X,Y,Z,W$ non-empty sets. Then, for any $xin X$, we have
$$((hcirc g)circ f)(x)=(hcirc g)(f(x))=h(g(f(x)))=h((gcirc f)(x))=(hcirc(gcirc f))(x)$$
and so $(hcirc g)circ f$ and $hcirc(gcirc f)$ are pointwise equal and thus equal as functions.
Note that this "proof" is really just analysis the nature of composition as we just shift the brackets around; there is no hidden meaning; composition is inherently associative.
An example would the space of all continuous functions from $mathbb R$ to $mathbb R$ which are invertible, or in other words bijective. Composition of continuous functions gives you a continuous function and the composition of two bijective functions is bijective.
Another classical example is the group of automorphism of an algebraic structure, e.g. for a vector space $V$ the set of all bijective linear maps $phi: Vto V$.
answered Mar 28 at 15:15
blubblub
2,770826
$endgroup$
$begingroup$
This is what I hoped and thought, but I did not remember any proof of associativity so I was not sure.
$endgroup$
– mathreadler
Mar 28 at 15:17
add a comment |
$begingroup$
You only need closure under composition and inverses as composition of functions is always associative.
You may prove this as follows: Let $f:Xto Y$, $g:Yto Z$, $h:Zto W$ be functions with $X,Y,Z,W$ non-empty sets. Then, for any $xin X$, we have
$$((hcirc g)circ f)(x)=(hcirc g)(f(x))=h(g(f(x)))=h((gcirc f)(x))=(hcirc(gcirc f))(x)$$
and so $(hcirc g)circ f$ and $hcirc(gcirc f)$ are pointwise equal and thus equal as functions.
Note that this "proof" is really just analysis the nature of composition as we just shift the brackets around; there is no hidden meaning; composition is inherently associative.
An example would the space of all continuous functions from $mathbb R$ to $mathbb R$ which are invertible, or in other words bijective. Composition of continuous functions gives you a continuous function and the composition of two bijective functions is bijective.
Another classical example is the group of automorphism of an algebraic structure, e.g. for a vector space $V$ the set of all bijective linear maps $phi: Vto V$.
answered Mar 28 at 15:15
blubblub
2,770826
$endgroup$
$begingroup$
This is what I hoped and thought, but I did not remember any proof of associativity so I was not sure.
$endgroup$
– mathreadler
Mar 28 at 15:17
add a comment |
$begingroup$
You only need closure under composition and inverses as composition of functions is always associative.
You may prove this as follows: Let $f:Xto Y$, $g:Yto Z$, $h:Zto W$ be functions with $X,Y,Z,W$ non-empty sets. Then, for any $xin X$, we have
$$((hcirc g)circ f)(x)=(hcirc g)(f(x))=h(g(f(x)))=h((gcirc f)(x))=(hcirc(gcirc f))(x)$$
and so $(hcirc g)circ f$ and $hcirc(gcirc f)$ are pointwise equal and thus equal as functions.
Note that this "proof" is really just analysis the nature of composition as we just shift the brackets around; there is no hidden meaning; composition is inherently associative.
An example would the space of all continuous functions from $mathbb R$ to $mathbb R$ which are invertible, or in other words bijective. Composition of continuous functions gives you a continuous function and the composition of two bijective functions is bijective.
Another classical example is the group of automorphism of an algebraic structure, e.g. for a vector space $V$ the set of all bijective linear maps $phi: Vto V$.
answered Mar 28 at 15:15
blubblub
2,770826
$endgroup$
You only need closure under composition and inverses as composition of functions is always associative.
You may prove this as follows: Let $f:Xto Y$, $g:Yto Z$, $h:Zto W$ be functions with $X,Y,Z,W$ non-empty sets. Then, for any $xin X$, we have
$$((hcirc g)circ f)(x)=(hcirc g)(f(x))=h(g(f(x)))=h((gcirc f)(x))=(hcirc(gcirc f))(x)$$
and so $(hcirc g)circ f$ and $hcirc(gcirc f)$ are pointwise equal and thus equal as functions.
Note that this "proof" is really just analysis the nature of composition as we just shift the brackets around; there is no hidden meaning; composition is inherently associative.
An example would the space of all continuous functions from $mathbb R$ to $mathbb R$ which are invertible, or in other words bijective. Composition of continuous functions gives you a continuous function and the composition of two bijective functions is bijective.
Another classical example is the group of automorphism of an algebraic structure, e.g. for a vector space $V$ the set of all bijective linear maps $phi: Vto V$.
answered Mar 28 at 15:15
blubblub
2,770826
edited Mar 28 at 15:23
answered Mar 28 at 15:15
blubblub
2,770826
answered Mar 28 at 15:15
blubblub
2,770826
answered Mar 28 at 15:15
blubblub
2,770826
2,770826
$begingroup$
This is what I hoped and thought, but I did not remember any proof of associativity so I was not sure.
$endgroup$
– mathreadler
Mar 28 at 15:17
add a comment |
$begingroup$
This is what I hoped and thought, but I did not remember any proof of associativity so I was not sure.
$endgroup$
– mathreadler
Mar 28 at 15:17
$begingroup$
This is what I hoped and thought, but I did not remember any proof of associativity so I was not sure.
$endgroup$
– mathreadler
Mar 28 at 15:17
$begingroup$
This is what I hoped and thought, but I did not remember any proof of associativity so I was not sure.
$endgroup$
– mathreadler
Mar 28 at 15:17
add a comment |
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3
$begingroup$
The functions would obviously need to be bijective, but other than that, this should hold.
$endgroup$
– Don Thousand
Mar 28 at 15:10
1
$begingroup$
I don't think polynomials are a good place to look since many are non-invertible (as many are not bijective).
$endgroup$
– Don Thousand
Mar 28 at 15:17
1
$begingroup$
Well, there's the group of order two consisting of the functions $x mapsto x$ and $x mapsto 1 - x$. If that's not exciting enough, and if one is allowed to compactify $mathbbR$ into $overlinemathbbR = mathbbR cup infty$, then one can enlarge that example into the real version of the anharmonic group, consisting of six functions $x mapsto x$, $x mapsto 1 - x$, $x mapsto frac1x$, $x mapsto frac11 - x$, $x mapsto 1 - frac1x$, and $x mapsto fracxx - 1$. I'll get me coat ...
$endgroup$
– Calum Gilhooley
Mar 28 at 20:13
1
$begingroup$
Perhaps more interestingly, and not requiring $infty$, there is the group of all non-constant affine functions, viz. $x mapsto ax + b$, where $a, b in mathbbR$ and $a ne 0$. And this has another subgroup, infinite this time, consisting of just the functions $x mapsto ax$, where $a ne 0$.
$endgroup$
– Calum Gilhooley
Mar 28 at 20:29
1
$begingroup$
But if $m(x)= cx^n$, then $m^-1(x) = sqrt[n]fracxc$, and not, as you write, $1/ccdot x^-n$.
$endgroup$
– Calum Gilhooley
Mar 28 at 21:49