6
$\begingroup$

Similarly to "center of a group", also "inner automorphism" has a topological sounding. But, though for the former I could find some possible explanation in this site, for the latter I couldn't. So:

Why are inner automorphisms named this way?

$\endgroup$
  • 1
    $\begingroup$ I wonder if the term "outer automorphism group" was coined first. This would make sense, as outer automorphism groups have been studied since at least the 1920s (e.g. Neilsen proved that the outer automorphism group of a surface group is the extended mapping class group of the surface), while on the other hand, inner automorphism groups are not studied, or at least not to the same extent. $\endgroup$ – user1729 May 30 at 12:57
3
$\begingroup$

If we consider the homomorphic image of $G$ in $\operatorname{Sym}(G)$ by left multiplication, say $L$, and the anti-homomorphic image of $G$ in $\operatorname{Sym}(G)$ by right multiplication, say $R$, then we have that $\operatorname{Inn}(G)= LR\cap\operatorname{Aut}(G)$ (incidentally, $L\cap R \cong Z(G)$). But $\operatorname{Aut}(G)\cap L=\operatorname{Aut}(G)\cap R=\{id\}$, and thence we can think of the non-indentity elements of $\operatorname{Aut}(G)$ as "spread away around" $L\cup R$ and overlapping $LR$ just in the "outermost" region of this latter, which is then the "innermost" of $\operatorname{Aut}(G)$ (deducted the singleton $\{id\}$). I do not claim that this is the hystorical reason for this name, that I don't know, but it is just my guess to make it sound reasonable.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Looks promising. I'm going to make a sketch of that. $\endgroup$ – csn Jun 1 at 11:31
9
$\begingroup$

My guess is that it's because they can be defined by a pretty natural operations from inside the group. The most basic operation we have is "multiply" (although "invert" is arguably 'more basic', but let me go on...). So we can look at a map like $$ f_a(x) = a \cdot x $$ which takes $G$ to $G$ (where $a$ is some fixed element of $G$). This isn't an automorphism, though, because it takes $e$ to $a$ rather than to itself. We could fix this by throwing in a $a^{-1}$, saying $$ g_a(x) = a^{-1} \cdot (a \cdot x) $$ but applying associativity reduces that to just the identity map, which is boring. We could, instead, fix the "$e$ goes to the wrong place" situation by multiplying by $a^{-1}$ on the other side: $$ h_a(x) = (a \cdot x)\cdot a^{-1}, $$ and that (at least for non-abelian groups) defines an automorphism, entirely using stuff from "inside $G$", so it's reasonable to call it an "inner automorphism."

To return to my original remark, "inversion" is pretty basic as well, so we could try $$ u(x) = x^{-1} $$ as a possible "internal automorphism". It certainly is "internal", and it even has $u(e) = e$, so we're ahead of the game. But when we look at $$ u(ab) = (ab)^{-1} = b^{-1}a^{-1} = u(b) u(a) $$ we see that unless the group is abelian, it's not a homorphism.

So for abelian groups, there's one "internally generated" automorphism (inversion) that jumps out at us, together with the identity automorphism. For non-abelian groups, there's a whole family of "internally generated" automorphisms (those defined via conjugation), among which is the identity (defined by conjugation by the identity element, for instance). Of course this "whole family" may contain only a few elements (i.e., many inner automorphisms may turn out to be the
same).

I don't know that this is where the term "inner" came from, but it sure seems reasonable.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I have seen them denoted as $\textrm{Int}(G)$, though I forget if this was standing for internal or intrinsic etc. Either way it emphasizes the viewpoint that they are automorphisms naturally obtained from $G$ itself. $\endgroup$ – zibadawa timmy May 31 at 3:51
1
$\begingroup$

Like the other answerers, I can only conjecture, but I would say this.

Given a group $G$, there is always some larger group $G'$ containing $G$ as a normal subgroup and having the property that every automorphism of $G$ is induced by conjugation by a suitable element of $a \in G'$. In that case, the automorphism of $G$ is "inner" if and only if $a$ can be selected to be in $G$ itself.

An example of such a group $G'$, in which all automorphisms of $G$ can be obtained by conjugation, would be the semidirect product $G \rtimes \text{Aut}(G)$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.