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Given an automorphism $f$ of a group $G$, is there some way to determine whether or not $f$ is inner? What are some techniques for doing this?
If $G$ is finite, this can clearly be done by brute force. This isn't a very insightful answer though, and not the sort of answer I'm hoping to get. Rather, I would like to acquire a better understanding of what properties distinguish inner automorphisms from other automorphisms.
I do not know a constructive way of checking whether an automorphism is inner or not, but there is the following theorem due to Schupp (see A characterization of inner automorphisms) which says the following :
An automorphism $\alpha$ of a group $G$ is inner iff whenever $G$ embeds in a group $H$, then $\alpha$ extends to an automorphism of $H$.
There appear to be some more special cases (for p-groups, for instance) where inner automorphisms have been characterized, but those statements are difficult to parse (for me)
Theorem: It is undecidable in general whether an automorphism is inner.
Proof: To see this, note that every group is the outer automorphism group of some group. Therefore, we can take some group $G$ such that $\operatorname{Out}(G)$ has insoluble word problem. In this group it is undecidable whether an automorphism $\alpha$ is in the kernel of the map $\operatorname{Aut}(G)\rightarrow\operatorname{Out}(G)$. This kernel is precisely the inner automorphisms, and so we are done.
Comment: We used the result that every group is the outer automorphism group of some group. See, for example, the paper of Bumagin-Wise entitled Every group is the outer automorphism group of a finitely generated group. Download it from Wise's website here (it is under "Every group is an outer automorphism group"). However, the Bumagin-Wise result does not give you a group $G$ which is finitely presentable (apart from in certain exceptional circumstances which are not applicable here). To see that there are finitely presented groups whose outer automorphism groups have insoluble word problem, begin by taking $A_0$ to be finitely presented with insoluble word problem and take $A:=A_0\ast \mathbb{Z}$. Then form $H=A\ast B$ where $B$ is finitely presented and non-trivial, and take the automorphism $\tau_a$ defined by $\tau_a(x)=a^{-1}xa$ for all $x\in A$ and $\tau_a(y)=y$ for all $y\in B$. Then $\tau_a$ is inner if and only if $a\in Z(A)$, but $Z(A)$ is trivial by construction. Thus, $\operatorname{Out}(H)$ has insoluble word problem, and the result is proven. This construction is due to Arzhantseva, Lafont and Minasyan.
Anyway, the following result is related to the theorem. I quite like it.
Lemma: It is undecidable in general whether two (inner) automorphisms define the same automorphism of the group.
Proof: To see this, take $G=H\ast\mathbb{Z}$ where $H$ has undecidable word problem (so $G$ has undecidable word problem and trivial centre). Write $\gamma_g$ and $\gamma_h$ for the inner automorphisms of $G$ corresponding to $g$ and $h$ respectively. Then $\gamma_g=\gamma_h$ if and only if $gh^{-1}=1$, which is undecidable as $G$ has undecidable word problem.
I will discuss that problem in finite groups.
First, note that the notion of an inner automorphism is not universal in the sense that any automorphism $u$ of a group $G$, can be considered as inner in a larger group (consider for instance the semi-direct product of $G$ and $u$).
Now working with a fixed group $G$, some of the propeties of the inner automorphisms (unfortunately, shared with other automorphisms) is that they act trivially on the center, they preserve conjugacy classes and act trivially on $G/G'$. However, these facts are in general so vague to decide whether an automorphism is inner or not.
A more powerful method is using cohomology (is some cases), for instance if our group contains a self centralizing normal subgroup $N$, one has an isomorphism between the group of $1$-cocycles $Z^1(G/N, Z(N))$ and the group of automorphisms acting trivially on $G/Z(N)$ and $N$, and this isomorphism maps the subgroup of $1$-cobords exactely to the group of inner automorphisms acting trivially on $G/Z(N)$ and $N$. So, the problem of having a non inner automorphism in this class amounts to proving that the cohomology group $H^1(G/N, Z(N))$ is not trivial. This interesting relation was used by Gaschutz in proving his famous result, that every finite non-simple p-group has outer automorphisms.
Finally, I note that there is many outstanding problems related to deciding that some automorphisms are inner or not, and the existing tools seem to be very unsatisfactory.
There is an interesting theorem of Gary Seitz and Walter Feit (1984): Let $G$ be a finite simple group and let $\alpha \in \operatorname{Aut}(G)$. Suppose that $\alpha[C] = C$ for all conjugacy classes $C$ of $G$. Then $\alpha \in \operatorname{Inn}(G)$. The proof uses the classification of finite simple groups ...
