Why are "inner" automorphisms named this way?

Question

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?

Answer 1

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.

Answer 2

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 $G \rtimes \text{Aut}(G)$

Answer 3

No more than a guess. Every $G$ realizes via left and right multiplication maps by a fixed $a\in G$, say $\lambda_a,\rho_a\in\operatorname{Sym}(G)$ respectively, with $a$ ranging throughout $G$. In a sense, $G$ as a group "is" the pair $(\Lambda,R)$, where $\Lambda:=\{\lambda_a, a\in G\}$ and $R:=\{\rho_a, a\in G\}$ are subgroups of $\operatorname{Sym}(G)$. Now, we can think of "$G$ in $\operatorname{Sym}(G)$" as the subgroup $\Lambda R(=R\Lambda)\le\operatorname{Sym}(G)$. It turns out that:

$$\operatorname{Aut}(G)\cap \Lambda R=\{\lambda_a\rho_{a^{-1}},a\in G\}\tag 1$$

Therefore, the RHS of $(1)$ comprises all and only the automorphisms of $G$ which lie inside "$G$ in $\operatorname{Sym}(G)$" (namely $\Lambda R$), which then can be rightly said "inner automorphisms".

Answer 4

I looked in some early sources and I think I have found the definitive answer. I'll give the short summary first, but it may be hard to appreciate without the historical context. Please suspend your disbelief while you read this; I promise to explain.

Suppose $\varphi$ is an automorphism of $G$. There always exists an element $s$ that allows us to write $\varphi$ in the form $g\mapsto s^{-1}gs$. Then $\varphi$ is called an inner or an outer automorphism according to whether $s$ is in $G$ or out of $G$.

Now I should explain the strange claim that such $s$ exists even for an outer automorphism. The claim is not actually incorrect, but only old-fashioned.

You need to understand that until at least the 1930s, the study of group theory was much less abstract than it is today. Typically, a group was not considered to be a set equipped with an arbitrary binary operation, etc. Rather:

Groups were understood to be groups of permutations.

The term “group” itself was understood as an abbreviation of the phrase “group of permutations”, and the crucial property was closure.

In the modern view, the idea of taking two completely arbitrary groups $G$ and $H$ and forming $G\cap H$ is nonsense, or at least confused, because the two group operations are not necessarily the same, and indeed $G$ and $H$ may not have the same identity element, or any elements in common. But up through the early 20th century, $G\cap H$ was perfectly clear: $G$ and $H$ have the same operation, because the operation is composition of permutations, and $G$ and $H$ have the same identity element, because in both cases it is the identity permutation.

(Viewed in this way, Cayley's theorem is a great deal more obvious than in a modern presentation!)

With that context established, here is the relevant paragraph from the earliest source I could find, Harold Hilton, An Introduction to the Theory of Groups of Finite Order, Clarendon Press 1908:

A simple isomorphism of $G$ with itself is often called an automorphism of $G$. It is defined as an inner or cogredient automorphism, if the isomorphism can be obtained by transforming $G$ by an element contained in $G$; if not as an outer or contragredient automorphism.

The key point is implicit, so it is easy to miss. But here it is: we can still understand an automorphism $\varphi$ as being a “transformation” of $G$ — that is, as a mapping $g \mapsto s^{-1}gs$ — even when $s$ is not an element of $G$.

To take the simplest possible example, consider $A_3 = \{(), (1\ 2\ 3), (1\ 3\ 2)\}$. There is an obvious nontrivial automorphism $\varphi$, which is clearly outer. In 2021, we would say that $\varphi$ does not have the form $g\mapsto s^{-1}gs$. But this is not quite true! It is true that there is no such $s$ in $A_3$. But if we remember to think like an early group theorist, we view $A_3$ as always being a subgroup of $S_3$. Now let $s = (2\ 3)$ and see that $\varphi$ does in fact map each $g$ to $s^{-1}gs$.

It's not hard to see that such an $s$ must exist for any automorphism of $G$, if one is prepared to construe $G$ as a subgroup of a larger symmetric group. (Cayley's theorem again.)