All Questions
8 questions
1
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1
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209
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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.
15
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1
answer
753
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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 ...
1
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0
answers
125
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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 $\operatorname{Hol}(Q_{2^{n+1}})$ ...
8
votes
2
answers
501
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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 ...
3
votes
1
answer
293
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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}...
9
votes
1
answer
3k
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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\...
3
votes
1
answer
1k
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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 $...
37
votes
5
answers
11k
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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 ...