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Obviously the automorphism groups of finite dimensional vector spaces are studied in great detail. By changing presentations of the vector space (by which I mean isomorphisms to some $K^{\oplus n}$) we can characterize automorphisms as matrices in $GL(n,K)$ and bring them in suitable shapes (thinking of Smith normal form, Jordan normal form etc).

Categorically this might be due to the very special fact that every vector space is free (in particular there are no relations between generators) and that finite products and finite coproducts coincide, making computing automorphisms rather easy.

But other than linear algebra I could not find other fields in mathematics, where automorphism groups are studied in that detail, which might be due to the presence of said relations. In fact it is common to reduce the study of some structure to linear algebra (e.g. via representation theory) to exploit matrices.

So I was wondering, whether there are other categories, in which automorphism groups are of similar use, while being somewhat accessible/computable. Let me give some examples, to clarify what I mean.

  • In Galois theory automorphisms of a field extension $L\mid K$ play a central role. I don’t know how easily they can be computed in specific instances though...
  • Do the classification results of linear algebra carry over to automorphisms of finitely generated abelian groups? It seems like there we can exploit the structure theorem to get at least a block diagonal matrix form. I was unable to find references...
  • Automorphism groups of simple graphs seem to be rather difficult to compute, so it doesn’t seem like there is a general theory of automorphism groups of simple graphs. I would guess there are certain subcategories of good graphs, where automorphisms are better behaved though...
Jonas Linssen
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    Are you asking specifically about groups? We can talk about automorphism groups of fields, rings, graphs, designs, geometric objects, etc. In each case, automorphism groups are well-studied, and can be said to be well-understood for many families of examples. Can you be specific about what exactly you are looking for here? – xxxxxxxxx Dec 26 '20 at 14:18
  • @MorganRidgers I am curious whether this is a typo or "design" is a name of an actual mathematical object? – lisyarus Dec 26 '20 at 14:52
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    Concerning finitely generated abelian groups specifically, you can start with this answer: https://math.stackexchange.com/questions/55262/the-automorphism-group-of-a-direct-product-of-abelian-groups-is-isomorphic-to-a – lisyarus Dec 26 '20 at 15:04
  • @MorganRodgers thank, you very much. I think the question was vague indeed, yet I don’t know whether my edit made it more specific... – Jonas Linssen Dec 26 '20 at 16:08
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    @lisyarus https://en.m.wikipedia.org/wiki/Combinatorial_design – Alex Kruckman Dec 26 '20 at 16:39
  • @AlexKruckman I see, thank you! – lisyarus Dec 26 '20 at 16:54
  • The group $\operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q)$ is the main point of study in the vast topic of algebraic number theory and still an active field of research. – Qi Zhu Dec 27 '20 at 13:00
  • Self-diffeomorphism groups of manifolds are extremely interesting and much studied objects. They're mostly accessible through their topological invariants though – Nobody Dec 29 '20 at 02:26

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Let's say we are talking about groups; though rings and fields, among others, are also interesting.

Due to Cayley's theorem, every group can be embedded in a symmetric group.

For $\mathfrak S_n$ (finite symmetric groups), other than $n=6$, all the automorphisms are inner.

If $n\ne2,6$, then $\rm{Aut}\mathfrak S_n\cong\mathfrak S_n$.

Also, due to the fact that there is only one way to extend an automorphism from $\mathfrak A_n$, the $n$-th alternating group, to $\mathfrak S_n$, we get that $\rm{Aut}\mathfrak S_n\cong\rm{Aut}\mathfrak A_n$.

Next, I would like to mention the automorphisms of finite cyclic groups, $\Bbb Z_n$. They are fairly easy to understand, since an automorphism has to take a generator to a generator. We easily get that $\rm{Aut}\Bbb Z_n\cong\Bbb Z_n^\times$.

The infinite cyclic group, $\Bbb Z$, on the other hand, has just two automorphisms, one the trivial one and one sending $1$ to $-1$.

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    So automorphisms of finite symmetric groups are nice, because they are just conjugation. This is lovely. Just out of curiosity: where does the exception 6 come from? – Jonas Linssen Dec 26 '20 at 12:30
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    Why does the automorphism group of G have anything to do with automorphism group of a group that contains G? – lisyarus Dec 26 '20 at 14:14
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    I agree with lisyarus, I don't see the relevance of the first line of this. – xxxxxxxxx Dec 26 '20 at 14:16
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    In particular, since vector spaces are abelian groups (wrt addition), their (group-theoretic) automorphism groups are at least as large as $\operatorname{GL}(n)$ coupled with automorphisms of the ground field, while the inner automorphism group is trivial (since the vector space is an abelian group). – lisyarus Dec 26 '20 at 14:19
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    @ChrisCuster Why this attitude? Downvotes are pretty normal and provide a mechanism to indicate that a post has certain problems. In this case, the sentence on the Cayley theorem seems to implicitly imply that the automorphism of a group has something to do with inner automorphisms of symmetric groups, which I demonstrated not to be the case in my previous comment. If I misunderstood your answer, I will happily retract my downvote after we resolve this misunderstanding. – lisyarus Dec 26 '20 at 14:47
  • I think that what Chris is trying to say by the first line is that since any group is isomorphic to some subgroup of a symmetric group, then it may come in handy to know things about the automorphism group of symmetric groups (anyway, I am not sure how this helps, maybe he can explain this further. I mean, our group $G$ is not going to be isomorphic to a symmetric group, but to a group of transformations, and this doesn't tell us anything about $\operatorname{Aut}(G)$, even if we know the automorphism group of those finite symmetric groups) – Alexdanut Dec 26 '20 at 21:44
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    @PrudiiArca: In a sense, it’s almost just a coincidence that there is one when $n=6$. It just so happens that the conjugacy class of transpositions in $S_6$ has the same number of elements as the conjugacy class of products of three disjoint transpositions, and that one can define an automorphism that sends transpositions to products of three disjoint transpositions. For other values of $n$, this does not happen, and there is nowhere to send transpositions except to transpositions, which ends up leading to inner automorphisms. It’s an instance of the “law of small numbers” (cont) – Arturo Magidin Dec 26 '20 at 23:45
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    @PrudiiArca: which informally says that there aren’t enough small numbers to avoid having coincidences happen. – Arturo Magidin Dec 26 '20 at 23:46
  • @ArturoMagidin Thank you! – Jonas Linssen Dec 27 '20 at 02:45