9

Let’s call a group $G$ invertible, if $\forall H \triangleleft G$, there $\exists K \triangleleft G$, such, that $K \cong \frac{G}{H}$ and $H \cong \frac{G}{K}$

All finite abelian groups are known to be invertible. All simple groups also are invertible.

However the vast majority of finite groups seems to be non-invertible: that includes symmetric groups, dihedral groups, holomorphs of cyclic groups, generalised quaternion groups and so on…

So my question is:

Is there some sort of classification of invertible finite groups?

Chain Markov
  • 16,012
  • 1
    Are the isomorphisms $K\to G/H$ and $H\to G/K$ allowed to be arbitrary or should they be induced by the inclusions? – Christoph Jul 14 '19 at 09:10
  • 1
    @Christoph, yes, they are allowed to be arbitrary. – Chain Markov Jul 14 '19 at 09:16
  • 4
    A direct product of a finite abelian group with a collection of nonabelian simple groups is also invertible. – Derek Holt Jul 14 '19 at 18:56
  • It almost seems your definition itself is a classification of sorts ... are there other ways of defining what an invertible group is besides the one you have given? –  Jul 14 '19 at 22:47
  • How is the definition of an invertible group close to a classification in itself? For example, can you name a few invertible $p$-groups of small order? – the_fox Jul 14 '19 at 22:57

0 Answers0