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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.

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  • $\begingroup$ do you min to share some ideas what ever you thought could probably led to the conclusion.... $\endgroup$ – user87543 Sep 15 '13 at 3:04
  • $\begingroup$ Inner automorphisms don't move elements outside their conjugacy classes. Mapping all the conjugay classes to themselves is probably not a sufficient condition for an automorphism to be inner (IDK), but it is certainly a necessary condition. $\endgroup$ – Jyrki Lahtonen Sep 15 '13 at 7:07
  • $\begingroup$ Also have a look at : mathoverflow.net/questions/36175/… $\endgroup$ – Prahlad Vaidyanathan Sep 15 '13 at 7:34
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    $\begingroup$ @JyrkiLahtonen, indeed not arxiv.org/pdf/1002.1359.pdf $\endgroup$ – Andreas Caranti Sep 15 '13 at 8:07
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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)

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  • $\begingroup$ This is extremely interesting and just the sort of thing I was looking for. Many thanks! $\endgroup$ – Alexander Sep 15 '13 at 14:08
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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.

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    $\begingroup$ This is very nice. $\endgroup$ – user43208 Sep 15 '13 at 12:27
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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.

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  • $\begingroup$ Many thanks. Another great answer! $\endgroup$ – Alexander Sep 16 '13 at 0:14
  • $\begingroup$ You welcome, good luck. $\endgroup$ – Yassine Guerboussa Sep 16 '13 at 11:30
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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 ...

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  • $\begingroup$ There are similar results in infinite group theory (most for hyperbolic-esque groups). A group $G$ is said to have Property A if $\alpha(g)$ is conjugate to $g$ for all $g\in G$ (this is equivalent to your conjugacy class condition). The definition is due to Grossman, who proved that if a group $G$ has Property A and is conjugacy separable then $\operatorname{Out}(G)$ is residually finite. Grossman used this to prove that mapping class groups (of closed, orientable 2-manifolds, if I recall correctly) are residually finite. Which was a big theorem at the time! $\endgroup$ – user1729 Sep 17 '13 at 12:47
  • $\begingroup$ Your remark is a very interesting addition indeed! I already was wondering if there is a larger class than the simple groups having Property A! $\endgroup$ – Nicky Hekster Sep 17 '13 at 12:56
  • $\begingroup$ If you are interested, I think the best paper to look up is this one of Metaftsis and Sykiotis (they make some nice observations). If you want to see where the theory can get you, look up this paper of Levitt and Minasyan. $\endgroup$ – user1729 Sep 17 '13 at 13:39
  • $\begingroup$ @ Nicky Hekster : I don't understand what you mean exactely, the result in your answer is that only the inner automorphisms preserve conjugacy classes, in a simple group. And the the property A mentioned by user1729, is that every automorphisms is class preserving. Thus, do you mean that a simple group has only inner automorphisms? $\endgroup$ – Yassine Guerboussa Sep 17 '13 at 15:04
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    $\begingroup$ Dear user1729, just for kidding, I have had a Professor who sometimes say, we are not in a tribunal; you have all the rights to do mistakes. Still, there is finite p-groups (far from being simple) in which all the class preserving automorphisms are inner. $\endgroup$ – Yassine Guerboussa Sep 17 '13 at 18:48

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