Are all verbal automorphisms inner power automorphisms?
$begingroup$
Suppose $G$ is a group. $DeclareMathOperator{Wa}{Wa}DeclareMathOperator{Tame}{Tame}DeclareMathOperator{Aut}{Aut}$
Lets call $phi in Aut(G)$ verbal automorphism iff $exists n in mathbb{N}, {a_i}_{i=0}^n subset G, {e_i}_{i=0}^n subset {-1; 1}$ such that $forall t in G$ $(phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n)$.
One can easily see, that all the verbal automorphisms form a normal subgroup in $Aut(G)$. Lets denote this subgroup as $Va(G)$. One can see, that sometimes $Va(G)$ is a proper subgroup: for example, $C_2 times C_2$ has no nontrivial verbal automorphisms, but $Aut(C_2 times C_2)$ is isomorphic to $S_3$. Also one can see that a subgroup is invariant under verbal automorphisms iff it is normal.
Lets call $phi in Aut(G)$ inner power automorphism iff it is a composition of an inner automorphism and a universal power automorphism. It is easy to see, that all inner power automorphisms form a normal subgroup in $Aut(G)$. Lets denote this subgroup as $Ip(G)$. One can also see that $Ip(G) leq Va(G)$ (as $forall phi in Ip(G) exists a in G, n in mathbb{Z}$ such that $forall t in G (phi(t) = a^{-1}t^na$)) and that $Ip(G)$ is always isomorphic to a homomorphic image of $frac{G}{Z(G)} times C_{exp(G)}$, where $exp(G)$ is the exponent of $G$.
Is the statement $Va(G) = Ip(G)$ always true?
If $G$ is abelian, then it definitely is, as all verbal automorphisms of any abelian group are universal power automorphisms. If $G$ is complete then the statement is also true, as all automorphisms of a complete group are inner. However, I failed to find out anything other than that.
Any help will be appreciated.
abstract-algebra group-theory normal-subgroups automorphism-group power-automorphism
asked Dec 15 '18 at 13:07
Yanior WegYanior Weg
1,77811138
$endgroup$
|
show 4 more comments
$begingroup$
Suppose $G$ is a group. $DeclareMathOperator{Wa}{Wa}DeclareMathOperator{Tame}{Tame}DeclareMathOperator{Aut}{Aut}$
Lets call $phi in Aut(G)$ verbal automorphism iff $exists n in mathbb{N}, {a_i}_{i=0}^n subset G, {e_i}_{i=0}^n subset {-1; 1}$ such that $forall t in G$ $(phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n)$.
One can easily see, that all the verbal automorphisms form a normal subgroup in $Aut(G)$. Lets denote this subgroup as $Va(G)$. One can see, that sometimes $Va(G)$ is a proper subgroup: for example, $C_2 times C_2$ has no nontrivial verbal automorphisms, but $Aut(C_2 times C_2)$ is isomorphic to $S_3$. Also one can see that a subgroup is invariant under verbal automorphisms iff it is normal.
Lets call $phi in Aut(G)$ inner power automorphism iff it is a composition of an inner automorphism and a universal power automorphism. It is easy to see, that all inner power automorphisms form a normal subgroup in $Aut(G)$. Lets denote this subgroup as $Ip(G)$. One can also see that $Ip(G) leq Va(G)$ (as $forall phi in Ip(G) exists a in G, n in mathbb{Z}$ such that $forall t in G (phi(t) = a^{-1}t^na$)) and that $Ip(G)$ is always isomorphic to a homomorphic image of $frac{G}{Z(G)} times C_{exp(G)}$, where $exp(G)$ is the exponent of $G$.
Is the statement $Va(G) = Ip(G)$ always true?
If $G$ is abelian, then it definitely is, as all verbal automorphisms of any abelian group are universal power automorphisms. If $G$ is complete then the statement is also true, as all automorphisms of a complete group are inner. However, I failed to find out anything other than that.
Any help will be appreciated.
abstract-algebra group-theory normal-subgroups automorphism-group power-automorphism
asked Dec 15 '18 at 13:07
Yanior WegYanior Weg
1,77811138
$endgroup$
2
$begingroup$
I am struggling to parse the statement "$exists n in mathbb{N} {a_i}_{i=0}^n subset G {e_i}_{i=0}^n subset {-1; 1} forall t in G phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n$." I am pretty sure I have worked it out, but it really isn't clear.
$endgroup$
– user1729
Dec 17 '18 at 15:03
$begingroup$
(Also, "tame" automorphisms was classically used to mean automorphisms of a group $G=langle Srangle$ induced by automorphisms of the ambiant free group $F(S)$.)
$endgroup$
– user1729
Dec 17 '18 at 15:04
$begingroup$
I think the following automorphism of the free group $F(a, b, c)$ is a counter-example: $phi: amapsto a^2b^{-1}$, $bmapsto bab^{-1}$, $cmapsto cb^2$. This automorphism certainly is not a word automorphism, but I don't have the time at the moment to think about the inner-automorphism bit.
$endgroup$
– user1729
Dec 17 '18 at 15:59
$begingroup$
(Also, there is a paper of Cohen, Metzler and Zimmerman, What does a basis of F(a,b) look like?, Math. Ann. (1981) which proves something very similar to what you are talking about, but for the free group on two generators.)
$endgroup$
– user1729
Dec 17 '18 at 16:01
$begingroup$
Is your definition of word automorphism missing commas? And a "such that" before the "for all"?
$endgroup$
– alex.jordan
Dec 17 '18 at 17:06
|
show 4 more comments
$begingroup$
Suppose $G$ is a group. $DeclareMathOperator{Wa}{Wa}DeclareMathOperator{Tame}{Tame}DeclareMathOperator{Aut}{Aut}$
Lets call $phi in Aut(G)$ verbal automorphism iff $exists n in mathbb{N}, {a_i}_{i=0}^n subset G, {e_i}_{i=0}^n subset {-1; 1}$ such that $forall t in G$ $(phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n)$.
One can easily see, that all the verbal automorphisms form a normal subgroup in $Aut(G)$. Lets denote this subgroup as $Va(G)$. One can see, that sometimes $Va(G)$ is a proper subgroup: for example, $C_2 times C_2$ has no nontrivial verbal automorphisms, but $Aut(C_2 times C_2)$ is isomorphic to $S_3$. Also one can see that a subgroup is invariant under verbal automorphisms iff it is normal.
Lets call $phi in Aut(G)$ inner power automorphism iff it is a composition of an inner automorphism and a universal power automorphism. It is easy to see, that all inner power automorphisms form a normal subgroup in $Aut(G)$. Lets denote this subgroup as $Ip(G)$. One can also see that $Ip(G) leq Va(G)$ (as $forall phi in Ip(G) exists a in G, n in mathbb{Z}$ such that $forall t in G (phi(t) = a^{-1}t^na$)) and that $Ip(G)$ is always isomorphic to a homomorphic image of $frac{G}{Z(G)} times C_{exp(G)}$, where $exp(G)$ is the exponent of $G$.
Is the statement $Va(G) = Ip(G)$ always true?
If $G$ is abelian, then it definitely is, as all verbal automorphisms of any abelian group are universal power automorphisms. If $G$ is complete then the statement is also true, as all automorphisms of a complete group are inner. However, I failed to find out anything other than that.
Any help will be appreciated.
abstract-algebra group-theory normal-subgroups automorphism-group power-automorphism
asked Dec 15 '18 at 13:07
Yanior WegYanior Weg
1,77811138
$endgroup$
Suppose $G$ is a group. $DeclareMathOperator{Wa}{Wa}DeclareMathOperator{Tame}{Tame}DeclareMathOperator{Aut}{Aut}$
Lets call $phi in Aut(G)$ verbal automorphism iff $exists n in mathbb{N}, {a_i}_{i=0}^n subset G, {e_i}_{i=0}^n subset {-1; 1}$ such that $forall t in G$ $(phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n)$.
One can easily see, that all the verbal automorphisms form a normal subgroup in $Aut(G)$. Lets denote this subgroup as $Va(G)$. One can see, that sometimes $Va(G)$ is a proper subgroup: for example, $C_2 times C_2$ has no nontrivial verbal automorphisms, but $Aut(C_2 times C_2)$ is isomorphic to $S_3$. Also one can see that a subgroup is invariant under verbal automorphisms iff it is normal.
Lets call $phi in Aut(G)$ inner power automorphism iff it is a composition of an inner automorphism and a universal power automorphism. It is easy to see, that all inner power automorphisms form a normal subgroup in $Aut(G)$. Lets denote this subgroup as $Ip(G)$. One can also see that $Ip(G) leq Va(G)$ (as $forall phi in Ip(G) exists a in G, n in mathbb{Z}$ such that $forall t in G (phi(t) = a^{-1}t^na$)) and that $Ip(G)$ is always isomorphic to a homomorphic image of $frac{G}{Z(G)} times C_{exp(G)}$, where $exp(G)$ is the exponent of $G$.
Is the statement $Va(G) = Ip(G)$ always true?
If $G$ is abelian, then it definitely is, as all verbal automorphisms of any abelian group are universal power automorphisms. If $G$ is complete then the statement is also true, as all automorphisms of a complete group are inner. However, I failed to find out anything other than that.
Any help will be appreciated.
abstract-algebra group-theory normal-subgroups automorphism-group power-automorphism
abstract-algebra group-theory normal-subgroups automorphism-group power-automorphism
asked Dec 15 '18 at 13:07
Yanior WegYanior Weg
1,77811138
asked Dec 15 '18 at 13:07
Yanior WegYanior Weg
1,77811138
edited Dec 19 '18 at 12:23
Yanior Weg
asked Dec 15 '18 at 13:07
Yanior WegYanior Weg
1,77811138
asked Dec 15 '18 at 13:07
Yanior WegYanior Weg
1,77811138
asked Dec 15 '18 at 13:07
Yanior WegYanior Weg
1,77811138
1,77811138
2
$begingroup$
I am struggling to parse the statement "$exists n in mathbb{N} {a_i}_{i=0}^n subset G {e_i}_{i=0}^n subset {-1; 1} forall t in G phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n$." I am pretty sure I have worked it out, but it really isn't clear.
$endgroup$
– user1729
Dec 17 '18 at 15:03
$begingroup$
(Also, "tame" automorphisms was classically used to mean automorphisms of a group $G=langle Srangle$ induced by automorphisms of the ambiant free group $F(S)$.)
$endgroup$
– user1729
Dec 17 '18 at 15:04
$begingroup$
I think the following automorphism of the free group $F(a, b, c)$ is a counter-example: $phi: amapsto a^2b^{-1}$, $bmapsto bab^{-1}$, $cmapsto cb^2$. This automorphism certainly is not a word automorphism, but I don't have the time at the moment to think about the inner-automorphism bit.
$endgroup$
– user1729
Dec 17 '18 at 15:59
$begingroup$
(Also, there is a paper of Cohen, Metzler and Zimmerman, What does a basis of F(a,b) look like?, Math. Ann. (1981) which proves something very similar to what you are talking about, but for the free group on two generators.)
$endgroup$
– user1729
Dec 17 '18 at 16:01
$begingroup$
Is your definition of word automorphism missing commas? And a "such that" before the "for all"?
$endgroup$
– alex.jordan
Dec 17 '18 at 17:06
|
show 4 more comments
2
$begingroup$
I am struggling to parse the statement "$exists n in mathbb{N} {a_i}_{i=0}^n subset G {e_i}_{i=0}^n subset {-1; 1} forall t in G phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n$." I am pretty sure I have worked it out, but it really isn't clear.
$endgroup$
– user1729
Dec 17 '18 at 15:03
$begingroup$
(Also, "tame" automorphisms was classically used to mean automorphisms of a group $G=langle Srangle$ induced by automorphisms of the ambiant free group $F(S)$.)
$endgroup$
– user1729
Dec 17 '18 at 15:04
$begingroup$
I think the following automorphism of the free group $F(a, b, c)$ is a counter-example: $phi: amapsto a^2b^{-1}$, $bmapsto bab^{-1}$, $cmapsto cb^2$. This automorphism certainly is not a word automorphism, but I don't have the time at the moment to think about the inner-automorphism bit.
$endgroup$
– user1729
Dec 17 '18 at 15:59
$begingroup$
(Also, there is a paper of Cohen, Metzler and Zimmerman, What does a basis of F(a,b) look like?, Math. Ann. (1981) which proves something very similar to what you are talking about, but for the free group on two generators.)
$endgroup$
– user1729
Dec 17 '18 at 16:01
$begingroup$
Is your definition of word automorphism missing commas? And a "such that" before the "for all"?
$endgroup$
– alex.jordan
Dec 17 '18 at 17:06
2
2
$begingroup$
I am struggling to parse the statement "$exists n in mathbb{N} {a_i}_{i=0}^n subset G {e_i}_{i=0}^n subset {-1; 1} forall t in G phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n$." I am pretty sure I have worked it out, but it really isn't clear.
$endgroup$
– user1729
Dec 17 '18 at 15:03
$begingroup$
I am struggling to parse the statement "$exists n in mathbb{N} {a_i}_{i=0}^n subset G {e_i}_{i=0}^n subset {-1; 1} forall t in G phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n$." I am pretty sure I have worked it out, but it really isn't clear.
$endgroup$
– user1729
Dec 17 '18 at 15:03
$begingroup$
(Also, "tame" automorphisms was classically used to mean automorphisms of a group $G=langle Srangle$ induced by automorphisms of the ambiant free group $F(S)$.)
$endgroup$
– user1729
Dec 17 '18 at 15:04
$begingroup$
(Also, "tame" automorphisms was classically used to mean automorphisms of a group $G=langle Srangle$ induced by automorphisms of the ambiant free group $F(S)$.)
$endgroup$
– user1729
Dec 17 '18 at 15:04
$begingroup$
I think the following automorphism of the free group $F(a, b, c)$ is a counter-example: $phi: amapsto a^2b^{-1}$, $bmapsto bab^{-1}$, $cmapsto cb^2$. This automorphism certainly is not a word automorphism, but I don't have the time at the moment to think about the inner-automorphism bit.
$endgroup$
– user1729
Dec 17 '18 at 15:59
$begingroup$
I think the following automorphism of the free group $F(a, b, c)$ is a counter-example: $phi: amapsto a^2b^{-1}$, $bmapsto bab^{-1}$, $cmapsto cb^2$. This automorphism certainly is not a word automorphism, but I don't have the time at the moment to think about the inner-automorphism bit.
$endgroup$
– user1729
Dec 17 '18 at 15:59
$begingroup$
(Also, there is a paper of Cohen, Metzler and Zimmerman, What does a basis of F(a,b) look like?, Math. Ann. (1981) which proves something very similar to what you are talking about, but for the free group on two generators.)
$endgroup$
– user1729
Dec 17 '18 at 16:01
$begingroup$
(Also, there is a paper of Cohen, Metzler and Zimmerman, What does a basis of F(a,b) look like?, Math. Ann. (1981) which proves something very similar to what you are talking about, but for the free group on two generators.)
$endgroup$
– user1729
Dec 17 '18 at 16:01
$begingroup$
Is your definition of word automorphism missing commas? And a "such that" before the "for all"?
$endgroup$
– alex.jordan
Dec 17 '18 at 17:06
$begingroup$
Is your definition of word automorphism missing commas? And a "such that" before the "for all"?
$endgroup$
– alex.jordan
Dec 17 '18 at 17:06
|
show 4 more comments
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$begingroup$
I am struggling to parse the statement "$exists n in mathbb{N} {a_i}_{i=0}^n subset G {e_i}_{i=0}^n subset {-1; 1} forall t in G phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n$." I am pretty sure I have worked it out, but it really isn't clear.
$endgroup$
– user1729
Dec 17 '18 at 15:03
$begingroup$
(Also, "tame" automorphisms was classically used to mean automorphisms of a group $G=langle Srangle$ induced by automorphisms of the ambiant free group $F(S)$.)
$endgroup$
– user1729
Dec 17 '18 at 15:04
$begingroup$
I think the following automorphism of the free group $F(a, b, c)$ is a counter-example: $phi: amapsto a^2b^{-1}$, $bmapsto bab^{-1}$, $cmapsto cb^2$. This automorphism certainly is not a word automorphism, but I don't have the time at the moment to think about the inner-automorphism bit.
$endgroup$
– user1729
Dec 17 '18 at 15:59
$begingroup$
(Also, there is a paper of Cohen, Metzler and Zimmerman, What does a basis of F(a,b) look like?, Math. Ann. (1981) which proves something very similar to what you are talking about, but for the free group on two generators.)
$endgroup$
– user1729
Dec 17 '18 at 16:01
$begingroup$
Is your definition of word automorphism missing commas? And a "such that" before the "for all"?
$endgroup$
– alex.jordan
Dec 17 '18 at 17:06