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37 votes
5 answers
11k views

When is Aut(G) abelian?

Let $G$ be a group such that $\operatorname{Aut}(G)$ is abelian. Is then $G$ abelian? This is a sort of generalization of the well-known exercise, that $G$ is abelian when $\operatorname{Aut}(G)$ is ...
Martin Brandenburg's user avatar
15 votes
1 answer
753 views

Finite abelian groups with fewer automorphisms than a subgroup

It is not too hard to find examples of finite groups which have fewer automorphisms than one of their subgroups. For example $\mathcal D_4 \times \mathbb Z/2\mathbb Z$ (where $\mathcal D_4$ is the ...
A. Bailleul's user avatar
  • 1,322
9 votes
1 answer
3k views

Automorphism group of a finite group

I would like to ask if there exists an explicit description of $\mathrm{Aut}(G)$, the group of automorphisms of a finite group $G$, in particular, when $G$ is abelian. E.g., if $G = \mathbb{Z}/m\...
Hair80's user avatar
  • 675
8 votes
2 answers
501 views

On $p$-groups with abelian automorphism group

Let $G$ be a $p$-group of order $p^{n}\geq p^{7}$ and its automorphism group is elementary abelian $p$-group. Then, it is clear that $G$ is nilpotent of class $2$. However, the converse is not true in ...
Nourddine Snanou's user avatar
3 votes
1 answer
1k views

Automorphisms of an infinite direct product of abelian groups

Let $G = \prod_p \mathbb{Z}/p\mathbb{Z}$, where $p$ ranges over all primes, considered as an abelian group. What does $\text{Aut}(G)$ (or even $\text{End}(G)$) look like? I know that that if we take $...
Iian Smythe's user avatar
  • 3,115
3 votes
1 answer
293 views

Abelian torsion-free group with $\mathbb{Z}_2\times\mathbb{Z}$ as automorphism group

Let $A$ be an abelian torsion-free group. If $A$ is isomorphic with the group of rational numbers whose denominators are powers of, say, $2$, then its automorphism group is isomorphic with $\mathbb{Z}...
Alex Doe's user avatar
  • 287
1 vote
1 answer
209 views

A question about automorphism group of abelian group

Does anyone know any references that describe automorphism group $\operatorname{Aut}(\mathbb R^n\times \mathbb T^m)$? I searched for a long time but couldn't find it.
free's user avatar
  • 71
1 vote
0 answers
125 views

Is the commutator of the holomorph of generalized quaternion group abelian?

Let $Q_{2^{n}} = \langle x, y \mathrel\vert x^{2^{n-1}}=y^4 = 1, x^{2^{n-2}}=y^2, y^{-1}xy = x^{-1} \rangle$ be the generalized quaternion group of order $2^{n}$. Let $\operatorname{Hol}(Q_{2^{n+1}})$ ...
bidermeyer's user avatar