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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 $\Gamma:=\lbrace \gamma_a \mid a\in G \rbrace \le \operatorname{Sym}(G)$, such that:

  • $G \cong \Theta$;
  • $G \cong \Gamma$;
  • $\theta_a\gamma_b=\gamma_b\theta_a, \forall a,b \in G$ and then $\Theta\Gamma=\Gamma\Theta \le \operatorname{Sym}(G)$;
  • $Z(G) \cong \Theta \cap \Gamma$;
  • $\Theta \unlhd \Theta\Gamma$ and $\Gamma \unlhd \Theta\Gamma$.

Besides, coniugacy establishes the subgroup $\Phi:=im_\varphi = \lbrace \varphi_{a} \mid a \in G\rbrace \le \operatorname{Aut}(G) \le \operatorname{Sym}(G)$. 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: $(\sigma^{-1}\varphi_a\sigma)(b)=\sigma^{-1}(\varphi_a(\sigma(b)))=\sigma^{-1}(a^{-1}\sigma(b)a)=$ $\sigma^{-1}(a^{-1})b\sigma^{-1}(a)$; call $\tau:=\sigma^{-1} \in \operatorname{Aut}(G)$, then $(\sigma^{-1}\varphi_a\sigma)(b)=\tau(a^{-1})b\tau(a)=\tau(a)^{-1}b\tau(a)=$ $\varphi_{\tau(a)}(b)$, so that $\sigma^{-1}\varphi_a\sigma=\varphi_{\sigma^{-1}(a)} \in \Phi$.

$\blacksquare$

REMARK. $\operatorname{Out}(G):=\operatorname{Aut}(G)/\Phi$ is the (factor) group of outer automorphisms of $G$.

Proposition 1. $\Phi \le \Theta\Gamma$.

Proof. By definition of $\varphi_a$, $\theta_b$ and $\gamma_c$, it is $\varphi_a=\theta_ {a^{-1}}\gamma_a$, and then $\Phi \subseteq \Theta\Gamma$.

$\blacksquare$

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. $\Phi = \Theta\Gamma \cap \operatorname{Aut}(G)$.

Proof. $\theta_a\gamma_b \in \operatorname{Aut}(G)$ iff $(\theta_a\gamma_b)(cd)=(\theta_a\gamma_b)(c)(\theta_a\gamma_b)(d)$ iff $acdb=acbadb$ iff $e=ba$ iff $\theta_a\gamma_b=\theta_{b^{-1}}\gamma_b$ iff $\theta_a\gamma_b \in \Phi$.

$\blacksquare$

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 $\Theta\Gamma$.

All what above, has brought me to envisage the following pictures of "limit" and "in between" situations:


enter image description here

enter image description here

enter image description here

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.

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put on hold as too broad by verret, Alexander Gruber 2 days ago

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Use, for example, $\operatorname{Sym}$ for $\operatorname{Sym}$. $\endgroup$ – Shaun Feb 2 at 12:14
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Also, the second and third diagrams are confusing me. The question shows you realize that $\Theta\cap \operatorname{Aut}(G) = id.$ and $\Gamma\cap \operatorname{Aut}(G) = id.$ 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
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    $\begingroup$ 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