Isomorph-automorphic subgroup: Difference between revisions

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Definition

Symbol-free definition

A subgroup of a group is termed isomorph-automorphic if given any other isomorphic subgroup, there is an automorphism of the whole group, mapping the subgroup isomorphically to the other one.

Definition with symbols

A subgroup $H$ of a group $G$ is termed isomorph-automorphic if whenever there exists a subgroup $K$ of $G$ such that $H$ and $K$ are isomorphic groups, there exists an automorphism $\sigma$ of $G$ such that $\sigma (H)=K$ .

Examples

In a finite cyclic group, every subgroup is isomorph-automorphic (in fact, every subgroup is isomorph-free: no two subgroups are isomorphic).

Similarly, in a finite elementary Abelian group, every subgroup is isomorph-automorphic. That's because given any two subspaces of a finite-dimensional vector space that have the same dimension, there is an automorphism of the whole space taking one to the other.

By Sylow's theorem, every Sylow subgroup is isomorph-automorphic.

On the other hand, many subgroups are not isomorph-automorphic:

In $\mathbb {Z}$ , the group of integers, any nontrivial subgroup is of the form $m\mathbb {Z}$ , $m\neq 0$ , hence is isomorphic to $\mathbb {Z}$ . However, there is clearly no automorphism of $\mathbb {Z}$ mapping it to a proper subgroup.
In the direct product $\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /4\mathbb {Z}$ , the direct factor $\mathbb {Z} /2\mathbb {Z}$ is isomorphic to the subgroup of multiples of 2 in the direct factor $\mathbb {Z} /4\mathbb {Z}$ . However, there is no automorphism taking the first to the second.
In the dihedral group of order eight, the center is of order two, and any subgroup generated by a reflection has order two, so they are isomorphic. However, there is no automorphism taking the center to a subgroup generated by a reflection.

Relation with other properties

Stronger properties

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 A [[subgroup]] <math>H</math> of a [[group]] <math>G</math> is termed '''isomorph-automorphic''' if whenever there exists a subgroup <math>K</math> of <math>G</math> such that <math>H</math> and <math>K</math> are [[isomorphic groups]], there exists an [[automorphism]] <math>\sigma</math> of <math>G</math> such that <math>\sigma(H) = K</math>.
+==Examples==
+In a finite [[cyclic group]], every subgroup is isomorph-automorphic (in fact, every subgroup is [[isomorph-free subgroup|isomorph-free]]: no two subgroups are isomorphic).
+Similarly, in a finite [[elementary Abelian group]], every subgroup is isomorph-automorphic. That's because given any two subspaces of a finite-dimensional vector space that have the same dimension, there is an automorphism of the whole space taking one to the other.
+By [[Sylow's theorem]], every Sylow subgroup is isomorph-automorphic.
+On the other hand, many subgroups are not isomorph-automorphic:
+# In <math>\mathbb{Z}</math>, the group of integers, any nontrivial subgroup is of the form <math>m\mathbb{Z}</math>, <math>m \ne 0</math>, hence is isomorphic to <math>\mathbb{Z}</math>. However, there is clearly no automorphism of <math>\mathbb{Z}</math> mapping it to a proper subgroup.
+# In the direct product <math>\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}</math>, the direct factor <math>\mathbb{Z}/2\mathbb{Z}</math> is isomorphic to the subgroup of multiples of 2 in the direct factor <math>\mathbb{Z}/4\mathbb{Z}</math>. However, there is no automorphism taking the first to the second.
+# In the [[dihedral group:D8|dihedral group of order eight]], the center is of order two, and any subgroup generated by a reflection has order two, so they are isomorphic. However, there is no automorphism taking the center to a subgroup generated by a reflection.
 ==Relation with other properties==