Mixing
Right action induced by free group homomorphism
$begingroup$
I was reading this proof by Steinberg of Nielsen-Schreier theorem and i had a doubt about proposition 1, that basically is an alternative universal property of free groups. It says:
Let $X$ be a set and $F$ a group equipped with a map $i: X rightarrow F$.
Then $F$ is a free group on $X$ (with respect to the mapping $i$) if and only if given any set $A$ and any map $σ: X rightarrow S_A$, there is a unique action of $F$ on $A$ such that:
$ai(x)=sigma(x)(a)$
Note that in the paper when he says actions he will always refer to a right action. In the paper he says that the right implication was obvious and i thought so, but later when i tried to check it i came up with nothing.
My idea was to use the universal property of free groups and then the inducted homomorphism:
$F$ is free so given the map $sigma$ there's a unique homomorphism $varphi: F rightarrow S_A$ such that $varphi(i(x))=sigma(x)$
Then i thought that the unique actions could be the action $*: A times F rightarrow A$ with $(a,w) rightarrow varphi(w)(a)$. The only problem with that is that it is a left action and not a right action.
To define the equivalent right actions that should be $(a,w) rightarrow varphi(w^{-1})(a)$ but this way the condition $ai(x)=sigma(x)(a)$ is not true anymore.
Any ideas to prove this statement?
abstract-algebra group-theory group-actions free-groups
edited Feb 2 at 23:26
Eric Wofsey
193k14221352
asked Feb 2 at 22:00
Andrea LicataAndrea Licata
261
$endgroup$
add a comment |
$begingroup$
I was reading this proof by Steinberg of Nielsen-Schreier theorem and i had a doubt about proposition 1, that basically is an alternative universal property of free groups. It says:
Let $X$ be a set and $F$ a group equipped with a map $i: X rightarrow F$.
Then $F$ is a free group on $X$ (with respect to the mapping $i$) if and only if given any set $A$ and any map $σ: X rightarrow S_A$, there is a unique action of $F$ on $A$ such that:
$ai(x)=sigma(x)(a)$
Note that in the paper when he says actions he will always refer to a right action. In the paper he says that the right implication was obvious and i thought so, but later when i tried to check it i came up with nothing.
My idea was to use the universal property of free groups and then the inducted homomorphism:
$F$ is free so given the map $sigma$ there's a unique homomorphism $varphi: F rightarrow S_A$ such that $varphi(i(x))=sigma(x)$
Then i thought that the unique actions could be the action $*: A times F rightarrow A$ with $(a,w) rightarrow varphi(w)(a)$. The only problem with that is that it is a left action and not a right action.
To define the equivalent right actions that should be $(a,w) rightarrow varphi(w^{-1})(a)$ but this way the condition $ai(x)=sigma(x)(a)$ is not true anymore.
Any ideas to prove this statement?
abstract-algebra group-theory group-actions free-groups
edited Feb 2 at 23:26
Eric Wofsey
193k14221352
asked Feb 2 at 22:00
Andrea LicataAndrea Licata
261
$endgroup$
add a comment |
$begingroup$
I was reading this proof by Steinberg of Nielsen-Schreier theorem and i had a doubt about proposition 1, that basically is an alternative universal property of free groups. It says:
Let $X$ be a set and $F$ a group equipped with a map $i: X rightarrow F$.
Then $F$ is a free group on $X$ (with respect to the mapping $i$) if and only if given any set $A$ and any map $σ: X rightarrow S_A$, there is a unique action of $F$ on $A$ such that:
$ai(x)=sigma(x)(a)$
Note that in the paper when he says actions he will always refer to a right action. In the paper he says that the right implication was obvious and i thought so, but later when i tried to check it i came up with nothing.
My idea was to use the universal property of free groups and then the inducted homomorphism:
$F$ is free so given the map $sigma$ there's a unique homomorphism $varphi: F rightarrow S_A$ such that $varphi(i(x))=sigma(x)$
Then i thought that the unique actions could be the action $*: A times F rightarrow A$ with $(a,w) rightarrow varphi(w)(a)$. The only problem with that is that it is a left action and not a right action.
To define the equivalent right actions that should be $(a,w) rightarrow varphi(w^{-1})(a)$ but this way the condition $ai(x)=sigma(x)(a)$ is not true anymore.
Any ideas to prove this statement?
abstract-algebra group-theory group-actions free-groups
edited Feb 2 at 23:26
Eric Wofsey
193k14221352
asked Feb 2 at 22:00
Andrea LicataAndrea Licata
261
$endgroup$
I was reading this proof by Steinberg of Nielsen-Schreier theorem and i had a doubt about proposition 1, that basically is an alternative universal property of free groups. It says:
Let $X$ be a set and $F$ a group equipped with a map $i: X rightarrow F$.
Then $F$ is a free group on $X$ (with respect to the mapping $i$) if and only if given any set $A$ and any map $σ: X rightarrow S_A$, there is a unique action of $F$ on $A$ such that:
$ai(x)=sigma(x)(a)$
Note that in the paper when he says actions he will always refer to a right action. In the paper he says that the right implication was obvious and i thought so, but later when i tried to check it i came up with nothing.
My idea was to use the universal property of free groups and then the inducted homomorphism:
$F$ is free so given the map $sigma$ there's a unique homomorphism $varphi: F rightarrow S_A$ such that $varphi(i(x))=sigma(x)$
Then i thought that the unique actions could be the action $*: A times F rightarrow A$ with $(a,w) rightarrow varphi(w)(a)$. The only problem with that is that it is a left action and not a right action.
To define the equivalent right actions that should be $(a,w) rightarrow varphi(w^{-1})(a)$ but this way the condition $ai(x)=sigma(x)(a)$ is not true anymore.
Any ideas to prove this statement?
abstract-algebra group-theory group-actions free-groups
abstract-algebra group-theory group-actions free-groups
edited Feb 2 at 23:26
Eric Wofsey
193k14221352
asked Feb 2 at 22:00
Andrea LicataAndrea Licata
261
edited Feb 2 at 23:26
Eric Wofsey
193k14221352
asked Feb 2 at 22:00
Andrea LicataAndrea Licata
261
edited Feb 2 at 23:26
Eric Wofsey
193k14221352
edited Feb 2 at 23:26
Eric Wofsey
193k14221352
edited Feb 2 at 23:26
Eric Wofsey
193k14221352
193k14221352
asked Feb 2 at 22:00
Andrea LicataAndrea Licata
261
asked Feb 2 at 22:00
Andrea LicataAndrea Licata
261
asked Feb 2 at 22:00
Andrea LicataAndrea Licata
261
261
add a comment |
add a comment |
1 Answer
1
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$begingroup$
You just have to use the action $(a,w)tovarphi(w^{-1})(a)$ for the unique homomorphism $varphi:Fto S_A$ such that $varphi(i(x))=sigma(x)^{color{red}{-1}}$.
Alternatively, I would guess that actually the author is using the reverse composition operation on $S_A$ (not the usual composition order of functions), so that a homomorphism to $S_A$ gives a right action rather than a left action.
answered Feb 2 at 23:26
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
How can i state that such homomorphism exists and that is unique? Is this equivalent to the "standard" universal proof?
$endgroup$
– Andrea Licata
Feb 3 at 1:36
$begingroup$
It is literally just an instance of the usual universal property, for the map $xmapsto sigma(x)^{-1}$.
$endgroup$
– Eric Wofsey
Feb 3 at 2:26
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
You just have to use the action $(a,w)tovarphi(w^{-1})(a)$ for the unique homomorphism $varphi:Fto S_A$ such that $varphi(i(x))=sigma(x)^{color{red}{-1}}$.
Alternatively, I would guess that actually the author is using the reverse composition operation on $S_A$ (not the usual composition order of functions), so that a homomorphism to $S_A$ gives a right action rather than a left action.
answered Feb 2 at 23:26
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
How can i state that such homomorphism exists and that is unique? Is this equivalent to the "standard" universal proof?
$endgroup$
– Andrea Licata
Feb 3 at 1:36
$begingroup$
It is literally just an instance of the usual universal property, for the map $xmapsto sigma(x)^{-1}$.
$endgroup$
– Eric Wofsey
Feb 3 at 2:26
add a comment |
$begingroup$
You just have to use the action $(a,w)tovarphi(w^{-1})(a)$ for the unique homomorphism $varphi:Fto S_A$ such that $varphi(i(x))=sigma(x)^{color{red}{-1}}$.
Alternatively, I would guess that actually the author is using the reverse composition operation on $S_A$ (not the usual composition order of functions), so that a homomorphism to $S_A$ gives a right action rather than a left action.
answered Feb 2 at 23:26
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
How can i state that such homomorphism exists and that is unique? Is this equivalent to the "standard" universal proof?
$endgroup$
– Andrea Licata
Feb 3 at 1:36
$begingroup$
It is literally just an instance of the usual universal property, for the map $xmapsto sigma(x)^{-1}$.
$endgroup$
– Eric Wofsey
Feb 3 at 2:26
add a comment |
$begingroup$
You just have to use the action $(a,w)tovarphi(w^{-1})(a)$ for the unique homomorphism $varphi:Fto S_A$ such that $varphi(i(x))=sigma(x)^{color{red}{-1}}$.
Alternatively, I would guess that actually the author is using the reverse composition operation on $S_A$ (not the usual composition order of functions), so that a homomorphism to $S_A$ gives a right action rather than a left action.
answered Feb 2 at 23:26
Eric WofseyEric Wofsey
193k14221352
$endgroup$
You just have to use the action $(a,w)tovarphi(w^{-1})(a)$ for the unique homomorphism $varphi:Fto S_A$ such that $varphi(i(x))=sigma(x)^{color{red}{-1}}$.
Alternatively, I would guess that actually the author is using the reverse composition operation on $S_A$ (not the usual composition order of functions), so that a homomorphism to $S_A$ gives a right action rather than a left action.
answered Feb 2 at 23:26
Eric WofseyEric Wofsey
193k14221352
answered Feb 2 at 23:26
Eric WofseyEric Wofsey
193k14221352
answered Feb 2 at 23:26
Eric WofseyEric Wofsey
193k14221352
answered Feb 2 at 23:26
Eric WofseyEric Wofsey
193k14221352
193k14221352
$begingroup$
How can i state that such homomorphism exists and that is unique? Is this equivalent to the "standard" universal proof?
$endgroup$
– Andrea Licata
Feb 3 at 1:36
$begingroup$
It is literally just an instance of the usual universal property, for the map $xmapsto sigma(x)^{-1}$.
$endgroup$
– Eric Wofsey
Feb 3 at 2:26
add a comment |
$begingroup$
How can i state that such homomorphism exists and that is unique? Is this equivalent to the "standard" universal proof?
$endgroup$
– Andrea Licata
Feb 3 at 1:36
$begingroup$
It is literally just an instance of the usual universal property, for the map $xmapsto sigma(x)^{-1}$.
$endgroup$
– Eric Wofsey
Feb 3 at 2:26
$begingroup$
How can i state that such homomorphism exists and that is unique? Is this equivalent to the "standard" universal proof?
$endgroup$
– Andrea Licata
Feb 3 at 1:36
$begingroup$
How can i state that such homomorphism exists and that is unique? Is this equivalent to the "standard" universal proof?
$endgroup$
– Andrea Licata
Feb 3 at 1:36
$begingroup$
It is literally just an instance of the usual universal property, for the map $xmapsto sigma(x)^{-1}$.
$endgroup$
– Eric Wofsey
Feb 3 at 2:26
$begingroup$
It is literally just an instance of the usual universal property, for the map $xmapsto sigma(x)^{-1}$.
$endgroup$
– Eric Wofsey
Feb 3 at 2:26
add a comment |
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