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 $\sigma^{-1}\varphi_a\sigma=\varphi_{\sigma^{-1}(a)} \in \Phi$.


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 $\Phi \subseteq \Theta\Gamma$.


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 $\theta_a=\theta_e=\iota_{\operatorname{Sym}(G)}$.


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 $\theta_a\gamma_b \in \Phi$.


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:

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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.


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

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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