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Full invariance does not satisfy intermediate subgroup condition

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This article gives the statement, and possibly proof, of a subgroup property (i.e., fully invariant subgroup) not satisfying a subgroup metaproperty (i.e., intermediate subgroup condition).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about fully invariant subgroup| Get more facts about intermediate subgroup condition|

Contents

Statement

Statement with symbols

It is possible to have groups H \le K \le G such that H is a fully invariant subgroup of G but H is not a fully invariant subgroup of K.

Related facts

Intermediate subgroup condition

Other related facts

Proof

Abelian example of prime-cube order

Let p be any prime. Consider the group:

G := \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z}.

Let H = \mho^1(G) = \{ (0,pa) \} be the set of pth powers in G and K = Ω1(G) = {(a,pb} be the set of elements of order 1 or p. Then H \le K \le G, and:

Example of the dihedral group

Further information: dihedral group:D8, subgroup structure of dihedral group:D8

Let G be the dihedral group of order eight:

G := \langle a,x \mid a^4 = x^2 = e, xax = a^{-1} \rangle.

Let H = \langle a^2 \rangle and K = \langle a^2, x \rangle. Then:

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