Mixing
under what assumptions will $phi_G$ be surjective
$begingroup$
Let $G$ be a group. For each $gin G$, define $L_g:Gto G$ by $L_g(h)=gh$. Define $phi_G : Gto Bi(G)$ by $phi_G(g)=L_g$ where $Bi(G)$ denotes the bijection of $Gto G$. Now under what assumptions will $phi_G$ be surjective.
My attempts : It's easy to show that $phi_G$ is injective. Now we want $phi_G$ be surjective. Then it will be bijective. So if G is finite , lets say $|G|=n$. Then $|Bi(G)|=n!$. And bijection needs $|G|=|Bi(G)|$, i.e $n=n!$. So $n=1$ or $n=2$. But what about if G is infinite. Any hints would be helpful.
abstract-algebra
asked Jan 31 at 13:40
Jaqen ChouJaqen Chou
487110
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group. For each $gin G$, define $L_g:Gto G$ by $L_g(h)=gh$. Define $phi_G : Gto Bi(G)$ by $phi_G(g)=L_g$ where $Bi(G)$ denotes the bijection of $Gto G$. Now under what assumptions will $phi_G$ be surjective.
My attempts : It's easy to show that $phi_G$ is injective. Now we want $phi_G$ be surjective. Then it will be bijective. So if G is finite , lets say $|G|=n$. Then $|Bi(G)|=n!$. And bijection needs $|G|=|Bi(G)|$, i.e $n=n!$. So $n=1$ or $n=2$. But what about if G is infinite. Any hints would be helpful.
abstract-algebra
asked Jan 31 at 13:40
Jaqen ChouJaqen Chou
487110
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group. For each $gin G$, define $L_g:Gto G$ by $L_g(h)=gh$. Define $phi_G : Gto Bi(G)$ by $phi_G(g)=L_g$ where $Bi(G)$ denotes the bijection of $Gto G$. Now under what assumptions will $phi_G$ be surjective.
My attempts : It's easy to show that $phi_G$ is injective. Now we want $phi_G$ be surjective. Then it will be bijective. So if G is finite , lets say $|G|=n$. Then $|Bi(G)|=n!$. And bijection needs $|G|=|Bi(G)|$, i.e $n=n!$. So $n=1$ or $n=2$. But what about if G is infinite. Any hints would be helpful.
abstract-algebra
asked Jan 31 at 13:40
Jaqen ChouJaqen Chou
487110
$endgroup$
Let $G$ be a group. For each $gin G$, define $L_g:Gto G$ by $L_g(h)=gh$. Define $phi_G : Gto Bi(G)$ by $phi_G(g)=L_g$ where $Bi(G)$ denotes the bijection of $Gto G$. Now under what assumptions will $phi_G$ be surjective.
My attempts : It's easy to show that $phi_G$ is injective. Now we want $phi_G$ be surjective. Then it will be bijective. So if G is finite , lets say $|G|=n$. Then $|Bi(G)|=n!$. And bijection needs $|G|=|Bi(G)|$, i.e $n=n!$. So $n=1$ or $n=2$. But what about if G is infinite. Any hints would be helpful.
abstract-algebra
abstract-algebra
asked Jan 31 at 13:40
Jaqen ChouJaqen Chou
487110
asked Jan 31 at 13:40
Jaqen ChouJaqen Chou
487110
asked Jan 31 at 13:40
Jaqen ChouJaqen Chou
487110
asked Jan 31 at 13:40
Jaqen ChouJaqen Chou
487110
asked Jan 31 at 13:40
Jaqen ChouJaqen Chou
487110
487110
add a comment |
add a comment |
2 Answers
2
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$begingroup$
If $G$ has at least three distinct elements $1,a,b$, then there exists a bijection $Gto G$ that swaps $aleftrightarrow b$ while keeping everything else (including $1$) fixed.
However, any left multiplication that fixes $1$ must be the identity.
answered Jan 31 at 13:42
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
Hagen von Eitzen has given an argument that works in general, but since you gave a nice cardinality-based argument in the case that $n$ is finite, I thought I should mention that your argument also works if $n$ is infinite!
The key point is that the set $operatorname{Bij}(A)$ of bijections of a set $A$ has greater cardinality than $A$ for any set $A$ of cardinality at least 3. See this other math.SE question. So there cannot exist a bijection from $A$ to $operatorname{Bij}(A)$.
answered Jan 31 at 13:54
Ben Blum-SmithBen Blum-Smith
10.2k23087
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
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$begingroup$
If $G$ has at least three distinct elements $1,a,b$, then there exists a bijection $Gto G$ that swaps $aleftrightarrow b$ while keeping everything else (including $1$) fixed.
However, any left multiplication that fixes $1$ must be the identity.
answered Jan 31 at 13:42
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
If $G$ has at least three distinct elements $1,a,b$, then there exists a bijection $Gto G$ that swaps $aleftrightarrow b$ while keeping everything else (including $1$) fixed.
However, any left multiplication that fixes $1$ must be the identity.
answered Jan 31 at 13:42
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
If $G$ has at least three distinct elements $1,a,b$, then there exists a bijection $Gto G$ that swaps $aleftrightarrow b$ while keeping everything else (including $1$) fixed.
However, any left multiplication that fixes $1$ must be the identity.
answered Jan 31 at 13:42
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
If $G$ has at least three distinct elements $1,a,b$, then there exists a bijection $Gto G$ that swaps $aleftrightarrow b$ while keeping everything else (including $1$) fixed.
However, any left multiplication that fixes $1$ must be the identity.
answered Jan 31 at 13:42
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jan 31 at 13:42
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jan 31 at 13:42
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jan 31 at 13:42
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
add a comment |
add a comment |
$begingroup$
Hagen von Eitzen has given an argument that works in general, but since you gave a nice cardinality-based argument in the case that $n$ is finite, I thought I should mention that your argument also works if $n$ is infinite!
The key point is that the set $operatorname{Bij}(A)$ of bijections of a set $A$ has greater cardinality than $A$ for any set $A$ of cardinality at least 3. See this other math.SE question. So there cannot exist a bijection from $A$ to $operatorname{Bij}(A)$.
answered Jan 31 at 13:54
Ben Blum-SmithBen Blum-Smith
10.2k23087
$endgroup$
add a comment |
$begingroup$
Hagen von Eitzen has given an argument that works in general, but since you gave a nice cardinality-based argument in the case that $n$ is finite, I thought I should mention that your argument also works if $n$ is infinite!
The key point is that the set $operatorname{Bij}(A)$ of bijections of a set $A$ has greater cardinality than $A$ for any set $A$ of cardinality at least 3. See this other math.SE question. So there cannot exist a bijection from $A$ to $operatorname{Bij}(A)$.
answered Jan 31 at 13:54
Ben Blum-SmithBen Blum-Smith
10.2k23087
$endgroup$
add a comment |
$begingroup$
Hagen von Eitzen has given an argument that works in general, but since you gave a nice cardinality-based argument in the case that $n$ is finite, I thought I should mention that your argument also works if $n$ is infinite!
The key point is that the set $operatorname{Bij}(A)$ of bijections of a set $A$ has greater cardinality than $A$ for any set $A$ of cardinality at least 3. See this other math.SE question. So there cannot exist a bijection from $A$ to $operatorname{Bij}(A)$.
answered Jan 31 at 13:54
Ben Blum-SmithBen Blum-Smith
10.2k23087
$endgroup$
Hagen von Eitzen has given an argument that works in general, but since you gave a nice cardinality-based argument in the case that $n$ is finite, I thought I should mention that your argument also works if $n$ is infinite!
The key point is that the set $operatorname{Bij}(A)$ of bijections of a set $A$ has greater cardinality than $A$ for any set $A$ of cardinality at least 3. See this other math.SE question. So there cannot exist a bijection from $A$ to $operatorname{Bij}(A)$.
answered Jan 31 at 13:54
Ben Blum-SmithBen Blum-Smith
10.2k23087
answered Jan 31 at 13:54
Ben Blum-SmithBen Blum-Smith
10.2k23087
answered Jan 31 at 13:54
Ben Blum-SmithBen Blum-Smith
10.2k23087
answered Jan 31 at 13:54
Ben Blum-SmithBen Blum-Smith
10.2k23087
10.2k23087
add a comment |
add a comment |
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