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Full invariance does not satisfy intermediate subgroup condition
From Groupprops
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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Contents |
Statement
Statement with symbols
It is possible to have groups
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
- Homomorph-containment satisfies intermediate subgroup condition
- Homomorph-containing implies intermediately fully invariant
- Characteristicity does not satisfy intermediate subgroup condition
Proof
Abelian example of prime-cube order
Let p be any prime. Consider the group:
.
Let
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
, and:
- H is fully invariant in G: This is on account of it being an agemo subgroup -- the image of a pth power under an endomorphism continues to be a pth power.
- H is not fully invariant in K: In fact, K is
where L = {(a,0)}, so K is an elementary abelian group of order p2. 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:
.
Let
and
. Then:
- H is fully invariant in G: In fact,
, 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,
where
, so K is a Klein four-group and it admits an automorphism interchanging H and L.