Mixing
Visual representations of groups (in their symmetric groups)
$begingroup$
Given a group $G$, left and right multiplications establish the subgroups $Theta:=lbrace theta_a mid a in G rbrace le operatorname{Sym}(G)$ and $Gamma:=lbrace gamma_a mid ain G rbrace le operatorname{Sym}(G)$, such that:
$G cong Theta$;
$G cong Gamma$;
$theta_agamma_b=gamma_btheta_a, forall a,b in G$ and then $ThetaGamma=GammaTheta le operatorname{Sym}(G)$;
$Z(G) cong Theta cap Gamma$;
$Theta unlhd ThetaGamma$ and $Gamma unlhd ThetaGamma$.
Besides, coniugacy establishes the subgroup $Phi:=im_varphi = lbrace varphi_{a} mid a in Grbrace le operatorname{Aut}(G) le operatorname{Sym}(G)$. It turns out that $ker_varphi=Z(G)$, whence $Phi cong G/Z(G)$ (fundamental homomorphism theorem), and finally:
$G$ abelian $Leftrightarrow Z(G)=G Leftrightarrow Phi= lbrace iota_{operatorname{Sym}(G)} rbrace$;
$G$ centerless ($Z(G)=lbrace e rbrace$) $Leftrightarrow Phi cong G$.
REMARK. $Phi$ is nothing else but the group of inner automorphisms of $G$, differently denoted by $operatorname{Inn}(G)$ or $mathscr{I}(G)$.
Proposition 0. $Phi unlhd operatorname{Aut}(G)$.
Proof. $forall a,b in G, forall sigma in operatorname{Aut}(G)$, we get: $(sigma^{-1}varphi_asigma)(b)=sigma^{-1}(varphi_a(sigma(b)))=sigma^{-1}(a^{-1}sigma(b)a)=$ $sigma^{-1}(a^{-1})bsigma^{-1}(a)$; call $tau:=sigma^{-1} in operatorname{Aut}(G)$, then $(sigma^{-1}varphi_asigma)(b)=tau(a^{-1})btau(a)=tau(a)^{-1}btau(a)=$ $varphi_{tau(a)}(b)$, so that $sigma^{-1}varphi_asigma=varphi_{sigma^{-1}(a)} in Phi$.
$blacksquare$
REMARK. $operatorname{Out}(G):=operatorname{Aut}(G)/Phi$ is the (factor) group of outer automorphisms of $G$.
Proposition 1. $Phi le ThetaGamma$.
Proof. By definition of $varphi_a$, $theta_b$ and $gamma_c$, it is $varphi_a=theta_ {a^{-1}}gamma_a$, and then $Phi subseteq ThetaGamma$.
$blacksquare$
Proposition 2. $Phi cap Theta = Phi cap Gamma = lbrace iota_{operatorname{Sym}(G)}rbrace$.
Proof. $varphi_a in Theta Leftrightarrow exists b in G mid varphi_a = theta_b Leftrightarrow varphi_a(c) = theta_b(c), forall c in G Leftrightarrow a^{-1}ca=bc, forall c in G Rightarrow$ (take $c=a$) $a=ba Rightarrow b=e Rightarrow varphi_a=theta_e=iota_{operatorname{Sym}(G)}$.
Equivalently, $theta_a$ is a homomorphism (and then an automorphism of $G$) iff $theta_a(bc)=theta_a(b)theta_a(c)$ iff $abc=abac$ iff $a=e$ iff $theta_a=theta_e=iota_{operatorname{Sym}(G)}$.
$blacksquare$
Proposition 3. $Phi = ThetaGamma cap operatorname{Aut}(G)$.
Proof. $theta_agamma_b in operatorname{Aut}(G)$ iff $(theta_agamma_b)(cd)=(theta_agamma_b)(c)(theta_agamma_b)(d)$ iff $acdb=acbadb$ iff $e=ba$ iff $theta_agamma_b=theta_{b^{-1}}gamma_b$ iff $theta_agamma_b in Phi$.
$blacksquare$
It seems to me that Proposition 3 makes the wording "inner automorphisms" plausible: they are precisely the only automorphisms that lie inside the "widest border of $G$ in $operatorname{Sym}(G)$", namely $ThetaGamma$.
All what above, has brought me to envisage the following pictures of "limit" and "in between" situations:
I haven't got a specific question to ask, but rather if you can see some other "nice feature" I could add, or amend, on the picture.
abstract-algebra group-theory symmetric-groups visualization
asked Feb 2 at 11:39
LucaLuca
4041110
$endgroup$
|
show 9 more comments
$begingroup$
Given a group $G$, left and right multiplications establish the subgroups $Theta:=lbrace theta_a mid a in G rbrace le operatorname{Sym}(G)$ and $Gamma:=lbrace gamma_a mid ain G rbrace le operatorname{Sym}(G)$, such that:
$G cong Theta$;
$G cong Gamma$;
$theta_agamma_b=gamma_btheta_a, forall a,b in G$ and then $ThetaGamma=GammaTheta le operatorname{Sym}(G)$;
$Z(G) cong Theta cap Gamma$;
$Theta unlhd ThetaGamma$ and $Gamma unlhd ThetaGamma$.
Besides, coniugacy establishes the subgroup $Phi:=im_varphi = lbrace varphi_{a} mid a in Grbrace le operatorname{Aut}(G) le operatorname{Sym}(G)$. It turns out that $ker_varphi=Z(G)$, whence $Phi cong G/Z(G)$ (fundamental homomorphism theorem), and finally:
$G$ abelian $Leftrightarrow Z(G)=G Leftrightarrow Phi= lbrace iota_{operatorname{Sym}(G)} rbrace$;
$G$ centerless ($Z(G)=lbrace e rbrace$) $Leftrightarrow Phi cong G$.
REMARK. $Phi$ is nothing else but the group of inner automorphisms of $G$, differently denoted by $operatorname{Inn}(G)$ or $mathscr{I}(G)$.
Proposition 0. $Phi unlhd operatorname{Aut}(G)$.
Proof. $forall a,b in G, forall sigma in operatorname{Aut}(G)$, we get: $(sigma^{-1}varphi_asigma)(b)=sigma^{-1}(varphi_a(sigma(b)))=sigma^{-1}(a^{-1}sigma(b)a)=$ $sigma^{-1}(a^{-1})bsigma^{-1}(a)$; call $tau:=sigma^{-1} in operatorname{Aut}(G)$, then $(sigma^{-1}varphi_asigma)(b)=tau(a^{-1})btau(a)=tau(a)^{-1}btau(a)=$ $varphi_{tau(a)}(b)$, so that $sigma^{-1}varphi_asigma=varphi_{sigma^{-1}(a)} in Phi$.
$blacksquare$
REMARK. $operatorname{Out}(G):=operatorname{Aut}(G)/Phi$ is the (factor) group of outer automorphisms of $G$.
Proposition 1. $Phi le ThetaGamma$.
Proof. By definition of $varphi_a$, $theta_b$ and $gamma_c$, it is $varphi_a=theta_ {a^{-1}}gamma_a$, and then $Phi subseteq ThetaGamma$.
$blacksquare$
Proposition 2. $Phi cap Theta = Phi cap Gamma = lbrace iota_{operatorname{Sym}(G)}rbrace$.
Proof. $varphi_a in Theta Leftrightarrow exists b in G mid varphi_a = theta_b Leftrightarrow varphi_a(c) = theta_b(c), forall c in G Leftrightarrow a^{-1}ca=bc, forall c in G Rightarrow$ (take $c=a$) $a=ba Rightarrow b=e Rightarrow varphi_a=theta_e=iota_{operatorname{Sym}(G)}$.
Equivalently, $theta_a$ is a homomorphism (and then an automorphism of $G$) iff $theta_a(bc)=theta_a(b)theta_a(c)$ iff $abc=abac$ iff $a=e$ iff $theta_a=theta_e=iota_{operatorname{Sym}(G)}$.
$blacksquare$
Proposition 3. $Phi = ThetaGamma cap operatorname{Aut}(G)$.
Proof. $theta_agamma_b in operatorname{Aut}(G)$ iff $(theta_agamma_b)(cd)=(theta_agamma_b)(c)(theta_agamma_b)(d)$ iff $acdb=acbadb$ iff $e=ba$ iff $theta_agamma_b=theta_{b^{-1}}gamma_b$ iff $theta_agamma_b in Phi$.
$blacksquare$
It seems to me that Proposition 3 makes the wording "inner automorphisms" plausible: they are precisely the only automorphisms that lie inside the "widest border of $G$ in $operatorname{Sym}(G)$", namely $ThetaGamma$.
All what above, has brought me to envisage the following pictures of "limit" and "in between" situations:
I haven't got a specific question to ask, but rather if you can see some other "nice feature" I could add, or amend, on the picture.
abstract-algebra group-theory symmetric-groups visualization
asked Feb 2 at 11:39
LucaLuca
4041110
$endgroup$
1
$begingroup$
Use, for example,$operatorname{Sym}$for $operatorname{Sym}$.
$endgroup$
– Shaun
Feb 2 at 12:14
1
$begingroup$
Something seems wrong with your "G not abelian" picture. You have Inn(G) sitting inside $Theta cap Gamma$ which isn't right.
$endgroup$
– Ted
Feb 2 at 17:25
1
$begingroup$
In the abelian situation, if $|G|>2$, it is never the case that $operatorname{Aut}(G) = operatorname{Inn}(G)$, as appears to be depicted in the first picture. Abelian groups of order $>2$ always have nontrivial automorphisms, but no nontrivial inner automorphisms.
$endgroup$
– Ben Blum-Smith
Feb 7 at 13:46
1
$begingroup$
Also, the second and third diagrams are confusing me. The question shows you realize that $Thetacap operatorname{Aut}(G) = id.$ and $Gammacap operatorname{Aut}(G) = id.$ and $operatorname{Aut}(G)cap Phi$, but the diagrams seem to me to make it appear that $Phi$ is not contained in $operatorname{Aut}(G)$ and meanwhile $Phi$ contains $Theta$ and $Gamma$, whereas it meets them only trivially as you show in proposition 2.
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:00
1
$begingroup$
Incidentally, what you are calling "maximally nonabelian" is ordinarily called centerless. There are other competing possible meanings for "maximally nonabelian", for example perfect: en.wikipedia.org/wiki/Perfect_group. Also, a group can be centerless but pretty close to being abelian in other respects, for example $S_3$ is centerless even though it is solvable height 2 and actually metacyclic (en.wikipedia.org/wiki/Metacyclic_group).
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:02
|
show 9 more comments
$begingroup$
Given a group $G$, left and right multiplications establish the subgroups $Theta:=lbrace theta_a mid a in G rbrace le operatorname{Sym}(G)$ and $Gamma:=lbrace gamma_a mid ain G rbrace le operatorname{Sym}(G)$, such that:
$G cong Theta$;
$G cong Gamma$;
$theta_agamma_b=gamma_btheta_a, forall a,b in G$ and then $ThetaGamma=GammaTheta le operatorname{Sym}(G)$;
$Z(G) cong Theta cap Gamma$;
$Theta unlhd ThetaGamma$ and $Gamma unlhd ThetaGamma$.
Besides, coniugacy establishes the subgroup $Phi:=im_varphi = lbrace varphi_{a} mid a in Grbrace le operatorname{Aut}(G) le operatorname{Sym}(G)$. It turns out that $ker_varphi=Z(G)$, whence $Phi cong G/Z(G)$ (fundamental homomorphism theorem), and finally:
$G$ abelian $Leftrightarrow Z(G)=G Leftrightarrow Phi= lbrace iota_{operatorname{Sym}(G)} rbrace$;
$G$ centerless ($Z(G)=lbrace e rbrace$) $Leftrightarrow Phi cong G$.
REMARK. $Phi$ is nothing else but the group of inner automorphisms of $G$, differently denoted by $operatorname{Inn}(G)$ or $mathscr{I}(G)$.
Proposition 0. $Phi unlhd operatorname{Aut}(G)$.
Proof. $forall a,b in G, forall sigma in operatorname{Aut}(G)$, we get: $(sigma^{-1}varphi_asigma)(b)=sigma^{-1}(varphi_a(sigma(b)))=sigma^{-1}(a^{-1}sigma(b)a)=$ $sigma^{-1}(a^{-1})bsigma^{-1}(a)$; call $tau:=sigma^{-1} in operatorname{Aut}(G)$, then $(sigma^{-1}varphi_asigma)(b)=tau(a^{-1})btau(a)=tau(a)^{-1}btau(a)=$ $varphi_{tau(a)}(b)$, so that $sigma^{-1}varphi_asigma=varphi_{sigma^{-1}(a)} in Phi$.
$blacksquare$
REMARK. $operatorname{Out}(G):=operatorname{Aut}(G)/Phi$ is the (factor) group of outer automorphisms of $G$.
Proposition 1. $Phi le ThetaGamma$.
Proof. By definition of $varphi_a$, $theta_b$ and $gamma_c$, it is $varphi_a=theta_ {a^{-1}}gamma_a$, and then $Phi subseteq ThetaGamma$.
$blacksquare$
Proposition 2. $Phi cap Theta = Phi cap Gamma = lbrace iota_{operatorname{Sym}(G)}rbrace$.
Proof. $varphi_a in Theta Leftrightarrow exists b in G mid varphi_a = theta_b Leftrightarrow varphi_a(c) = theta_b(c), forall c in G Leftrightarrow a^{-1}ca=bc, forall c in G Rightarrow$ (take $c=a$) $a=ba Rightarrow b=e Rightarrow varphi_a=theta_e=iota_{operatorname{Sym}(G)}$.
Equivalently, $theta_a$ is a homomorphism (and then an automorphism of $G$) iff $theta_a(bc)=theta_a(b)theta_a(c)$ iff $abc=abac$ iff $a=e$ iff $theta_a=theta_e=iota_{operatorname{Sym}(G)}$.
$blacksquare$
Proposition 3. $Phi = ThetaGamma cap operatorname{Aut}(G)$.
Proof. $theta_agamma_b in operatorname{Aut}(G)$ iff $(theta_agamma_b)(cd)=(theta_agamma_b)(c)(theta_agamma_b)(d)$ iff $acdb=acbadb$ iff $e=ba$ iff $theta_agamma_b=theta_{b^{-1}}gamma_b$ iff $theta_agamma_b in Phi$.
$blacksquare$
It seems to me that Proposition 3 makes the wording "inner automorphisms" plausible: they are precisely the only automorphisms that lie inside the "widest border of $G$ in $operatorname{Sym}(G)$", namely $ThetaGamma$.
All what above, has brought me to envisage the following pictures of "limit" and "in between" situations:
I haven't got a specific question to ask, but rather if you can see some other "nice feature" I could add, or amend, on the picture.
abstract-algebra group-theory symmetric-groups visualization
asked Feb 2 at 11:39
LucaLuca
4041110
$endgroup$
Given a group $G$, left and right multiplications establish the subgroups $Theta:=lbrace theta_a mid a in G rbrace le operatorname{Sym}(G)$ and $Gamma:=lbrace gamma_a mid ain G rbrace le operatorname{Sym}(G)$, such that:
$G cong Theta$;
$G cong Gamma$;
$theta_agamma_b=gamma_btheta_a, forall a,b in G$ and then $ThetaGamma=GammaTheta le operatorname{Sym}(G)$;
$Z(G) cong Theta cap Gamma$;
$Theta unlhd ThetaGamma$ and $Gamma unlhd ThetaGamma$.
Besides, coniugacy establishes the subgroup $Phi:=im_varphi = lbrace varphi_{a} mid a in Grbrace le operatorname{Aut}(G) le operatorname{Sym}(G)$. It turns out that $ker_varphi=Z(G)$, whence $Phi cong G/Z(G)$ (fundamental homomorphism theorem), and finally:
$G$ abelian $Leftrightarrow Z(G)=G Leftrightarrow Phi= lbrace iota_{operatorname{Sym}(G)} rbrace$;
$G$ centerless ($Z(G)=lbrace e rbrace$) $Leftrightarrow Phi cong G$.
REMARK. $Phi$ is nothing else but the group of inner automorphisms of $G$, differently denoted by $operatorname{Inn}(G)$ or $mathscr{I}(G)$.
Proposition 0. $Phi unlhd operatorname{Aut}(G)$.
Proof. $forall a,b in G, forall sigma in operatorname{Aut}(G)$, we get: $(sigma^{-1}varphi_asigma)(b)=sigma^{-1}(varphi_a(sigma(b)))=sigma^{-1}(a^{-1}sigma(b)a)=$ $sigma^{-1}(a^{-1})bsigma^{-1}(a)$; call $tau:=sigma^{-1} in operatorname{Aut}(G)$, then $(sigma^{-1}varphi_asigma)(b)=tau(a^{-1})btau(a)=tau(a)^{-1}btau(a)=$ $varphi_{tau(a)}(b)$, so that $sigma^{-1}varphi_asigma=varphi_{sigma^{-1}(a)} in Phi$.
$blacksquare$
REMARK. $operatorname{Out}(G):=operatorname{Aut}(G)/Phi$ is the (factor) group of outer automorphisms of $G$.
Proposition 1. $Phi le ThetaGamma$.
Proof. By definition of $varphi_a$, $theta_b$ and $gamma_c$, it is $varphi_a=theta_ {a^{-1}}gamma_a$, and then $Phi subseteq ThetaGamma$.
$blacksquare$
Proposition 2. $Phi cap Theta = Phi cap Gamma = lbrace iota_{operatorname{Sym}(G)}rbrace$.
Proof. $varphi_a in Theta Leftrightarrow exists b in G mid varphi_a = theta_b Leftrightarrow varphi_a(c) = theta_b(c), forall c in G Leftrightarrow a^{-1}ca=bc, forall c in G Rightarrow$ (take $c=a$) $a=ba Rightarrow b=e Rightarrow varphi_a=theta_e=iota_{operatorname{Sym}(G)}$.
Equivalently, $theta_a$ is a homomorphism (and then an automorphism of $G$) iff $theta_a(bc)=theta_a(b)theta_a(c)$ iff $abc=abac$ iff $a=e$ iff $theta_a=theta_e=iota_{operatorname{Sym}(G)}$.
$blacksquare$
Proposition 3. $Phi = ThetaGamma cap operatorname{Aut}(G)$.
Proof. $theta_agamma_b in operatorname{Aut}(G)$ iff $(theta_agamma_b)(cd)=(theta_agamma_b)(c)(theta_agamma_b)(d)$ iff $acdb=acbadb$ iff $e=ba$ iff $theta_agamma_b=theta_{b^{-1}}gamma_b$ iff $theta_agamma_b in Phi$.
$blacksquare$
It seems to me that Proposition 3 makes the wording "inner automorphisms" plausible: they are precisely the only automorphisms that lie inside the "widest border of $G$ in $operatorname{Sym}(G)$", namely $ThetaGamma$.
All what above, has brought me to envisage the following pictures of "limit" and "in between" situations:
I haven't got a specific question to ask, but rather if you can see some other "nice feature" I could add, or amend, on the picture.
abstract-algebra group-theory symmetric-groups visualization
abstract-algebra group-theory symmetric-groups visualization
asked Feb 2 at 11:39
LucaLuca
4041110
asked Feb 2 at 11:39
LucaLuca
4041110
edited Feb 9 at 9:12
Luca
asked Feb 2 at 11:39
LucaLuca
4041110
asked Feb 2 at 11:39
LucaLuca
4041110
asked Feb 2 at 11:39
LucaLuca
4041110
4041110
1
$begingroup$
Use, for example,$operatorname{Sym}$for $operatorname{Sym}$.
$endgroup$
– Shaun
Feb 2 at 12:14
1
$begingroup$
Something seems wrong with your "G not abelian" picture. You have Inn(G) sitting inside $Theta cap Gamma$ which isn't right.
$endgroup$
– Ted
Feb 2 at 17:25
1
$begingroup$
In the abelian situation, if $|G|>2$, it is never the case that $operatorname{Aut}(G) = operatorname{Inn}(G)$, as appears to be depicted in the first picture. Abelian groups of order $>2$ always have nontrivial automorphisms, but no nontrivial inner automorphisms.
$endgroup$
– Ben Blum-Smith
Feb 7 at 13:46
1
$begingroup$
Also, the second and third diagrams are confusing me. The question shows you realize that $Thetacap operatorname{Aut}(G) = id.$ and $Gammacap operatorname{Aut}(G) = id.$ and $operatorname{Aut}(G)cap Phi$, but the diagrams seem to me to make it appear that $Phi$ is not contained in $operatorname{Aut}(G)$ and meanwhile $Phi$ contains $Theta$ and $Gamma$, whereas it meets them only trivially as you show in proposition 2.
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:00
1
$begingroup$
Incidentally, what you are calling "maximally nonabelian" is ordinarily called centerless. There are other competing possible meanings for "maximally nonabelian", for example perfect: en.wikipedia.org/wiki/Perfect_group. Also, a group can be centerless but pretty close to being abelian in other respects, for example $S_3$ is centerless even though it is solvable height 2 and actually metacyclic (en.wikipedia.org/wiki/Metacyclic_group).
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:02
|
show 9 more comments
1
$begingroup$
Use, for example,$operatorname{Sym}$for $operatorname{Sym}$.
$endgroup$
– Shaun
Feb 2 at 12:14
1
$begingroup$
Something seems wrong with your "G not abelian" picture. You have Inn(G) sitting inside $Theta cap Gamma$ which isn't right.
$endgroup$
– Ted
Feb 2 at 17:25
1
$begingroup$
In the abelian situation, if $|G|>2$, it is never the case that $operatorname{Aut}(G) = operatorname{Inn}(G)$, as appears to be depicted in the first picture. Abelian groups of order $>2$ always have nontrivial automorphisms, but no nontrivial inner automorphisms.
$endgroup$
– Ben Blum-Smith
Feb 7 at 13:46
1
$begingroup$
Also, the second and third diagrams are confusing me. The question shows you realize that $Thetacap operatorname{Aut}(G) = id.$ and $Gammacap operatorname{Aut}(G) = id.$ and $operatorname{Aut}(G)cap Phi$, but the diagrams seem to me to make it appear that $Phi$ is not contained in $operatorname{Aut}(G)$ and meanwhile $Phi$ contains $Theta$ and $Gamma$, whereas it meets them only trivially as you show in proposition 2.
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:00
1
$begingroup$
Incidentally, what you are calling "maximally nonabelian" is ordinarily called centerless. There are other competing possible meanings for "maximally nonabelian", for example perfect: en.wikipedia.org/wiki/Perfect_group. Also, a group can be centerless but pretty close to being abelian in other respects, for example $S_3$ is centerless even though it is solvable height 2 and actually metacyclic (en.wikipedia.org/wiki/Metacyclic_group).
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:02
1
1
$begingroup$
Use, for example,
$operatorname{Sym}$ for $operatorname{Sym}$.$endgroup$
– Shaun
Feb 2 at 12:14
$begingroup$
Use, for example,
$operatorname{Sym}$ for $operatorname{Sym}$.$endgroup$
– Shaun
Feb 2 at 12:14
1
1
$begingroup$
Something seems wrong with your "G not abelian" picture. You have Inn(G) sitting inside $Theta cap Gamma$ which isn't right.
$endgroup$
– Ted
Feb 2 at 17:25
$begingroup$
Something seems wrong with your "G not abelian" picture. You have Inn(G) sitting inside $Theta cap Gamma$ which isn't right.
$endgroup$
– Ted
Feb 2 at 17:25
1
1
$begingroup$
In the abelian situation, if $|G|>2$, it is never the case that $operatorname{Aut}(G) = operatorname{Inn}(G)$, as appears to be depicted in the first picture. Abelian groups of order $>2$ always have nontrivial automorphisms, but no nontrivial inner automorphisms.
$endgroup$
– Ben Blum-Smith
Feb 7 at 13:46
$begingroup$
In the abelian situation, if $|G|>2$, it is never the case that $operatorname{Aut}(G) = operatorname{Inn}(G)$, as appears to be depicted in the first picture. Abelian groups of order $>2$ always have nontrivial automorphisms, but no nontrivial inner automorphisms.
$endgroup$
– Ben Blum-Smith
Feb 7 at 13:46
1
1
$begingroup$
Also, the second and third diagrams are confusing me. The question shows you realize that $Thetacap operatorname{Aut}(G) = id.$ and $Gammacap operatorname{Aut}(G) = id.$ and $operatorname{Aut}(G)cap Phi$, but the diagrams seem to me to make it appear that $Phi$ is not contained in $operatorname{Aut}(G)$ and meanwhile $Phi$ contains $Theta$ and $Gamma$, whereas it meets them only trivially as you show in proposition 2.
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:00
$begingroup$
Also, the second and third diagrams are confusing me. The question shows you realize that $Thetacap operatorname{Aut}(G) = id.$ and $Gammacap operatorname{Aut}(G) = id.$ and $operatorname{Aut}(G)cap Phi$, but the diagrams seem to me to make it appear that $Phi$ is not contained in $operatorname{Aut}(G)$ and meanwhile $Phi$ contains $Theta$ and $Gamma$, whereas it meets them only trivially as you show in proposition 2.
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:00
1
1
$begingroup$
Incidentally, what you are calling "maximally nonabelian" is ordinarily called centerless. There are other competing possible meanings for "maximally nonabelian", for example perfect: en.wikipedia.org/wiki/Perfect_group. Also, a group can be centerless but pretty close to being abelian in other respects, for example $S_3$ is centerless even though it is solvable height 2 and actually metacyclic (en.wikipedia.org/wiki/Metacyclic_group).
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:02
$begingroup$
Incidentally, what you are calling "maximally nonabelian" is ordinarily called centerless. There are other competing possible meanings for "maximally nonabelian", for example perfect: en.wikipedia.org/wiki/Perfect_group. Also, a group can be centerless but pretty close to being abelian in other respects, for example $S_3$ is centerless even though it is solvable height 2 and actually metacyclic (en.wikipedia.org/wiki/Metacyclic_group).
$endgroup$
– Ben Blum-Smith
Feb 7 at 14:02
|
show 9 more comments
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1
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Use, for example,
$operatorname{Sym}$for $operatorname{Sym}$.$endgroup$
– Shaun
Feb 2 at 12:14
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Something seems wrong with your "G not abelian" picture. You have Inn(G) sitting inside $Theta cap Gamma$ which isn't right.
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– Ted
Feb 2 at 17:25
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In the abelian situation, if $|G|>2$, it is never the case that $operatorname{Aut}(G) = operatorname{Inn}(G)$, as appears to be depicted in the first picture. Abelian groups of order $>2$ always have nontrivial automorphisms, but no nontrivial inner automorphisms.
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– Ben Blum-Smith
Feb 7 at 13:46
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Also, the second and third diagrams are confusing me. The question shows you realize that $Thetacap operatorname{Aut}(G) = id.$ and $Gammacap operatorname{Aut}(G) = id.$ and $operatorname{Aut}(G)cap Phi$, but the diagrams seem to me to make it appear that $Phi$ is not contained in $operatorname{Aut}(G)$ and meanwhile $Phi$ contains $Theta$ and $Gamma$, whereas it meets them only trivially as you show in proposition 2.
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– Ben Blum-Smith
Feb 7 at 14:00
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Incidentally, what you are calling "maximally nonabelian" is ordinarily called centerless. There are other competing possible meanings for "maximally nonabelian", for example perfect: en.wikipedia.org/wiki/Perfect_group. Also, a group can be centerless but pretty close to being abelian in other respects, for example $S_3$ is centerless even though it is solvable height 2 and actually metacyclic (en.wikipedia.org/wiki/Metacyclic_group).
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– Ben Blum-Smith
Feb 7 at 14:02