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).
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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


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 p^{th} powers in G and K = \Omega_1(G) = \{(a,pb \} be the set of elements of order 1 or p. Then H \le K \le G, and:

  • H is fully invariant in G: This is on account of it being an agemo subgroup -- the image of a p^{th} power under an endomorphism continues to be a p^{th} power.
  • H is not fully invariant in K: In fact, K is L \times H where L = \{ (a,0) \}, so K is an elementary abelian group of order p^2. In particular, K has an automorphism interchanging the coordinates, and H is not invariant under this automorphism.

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:

  • H is fully invariant in G: In fact, H = \mho^1(G), and also H is the commutator subgroup of G, so H is fully invariant in G.
  • H is not fully invariant in K: In fact, K = H \times L where L = \langle x \rangle, so K is a Klein four-group and it admits an automorphism interchanging H and L.