Transitive normality is not finite-join-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., transitively normal subgroup) not satisfying a subgroup metaproperty (i.e., finite-join-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Strongly finite-join-closed subgroup property (?), .
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Statement

Verbal statement

The join of two transitively normal subgroups of a group need not be transitively normal.

Statement with symbols

Suppose G is a group and H, K are two transitively normal subgroups of G. Then, the join \langle H, K \rangle, which in this case equals the product HK, need not be a transitively normal subgroup.

Related facts

Related metaproperty dissatisfactions for transitively normal subgroups

Related properties that are join-closed

  • Normality is strongly join-closed: In particular, this implies that any join of transitively normal subgroups is a normal subgroup, even though it need not be transitively normal.

Proof

Further information: symmetric group:S3

Let A be the cyclic group of order three. Let B be the symmetric group of degree three, and C be the subgroup of order three in B. Define:

G = A \times B, \qquad H = A \times 1, \qquad K = 1 \times C.

Then, we have:

  • H and K are both normal subgroups of prime order. In particular, the only normal subgroups they have are the whole group and the trivial subgroup, both of which are normal in G. Thus, H and K are both transitively normal.
  • HK is not transitively normal: HK = A \times C. HK is elementary abelian of order nine. Pick an isomorphism \sigma:A \to C and consider the subgroup L of HK given by L = \{ (a, \sigma(a) \mid a \in A \}. Then, L is normal in HK, since HK is abelian. However, L is not normal in G, because conjugation by an element in 1 \times (B \setminus C) sends (a,\sigma(a)) to (a,\sigma(a)^2), which is not in L. Thus, HK is not transitively normal.

Note that in this example, H is the center of G and is also a direct factor of G. This shows that taking the join of a transitively normal subgroup with the center or with a direct factor does not guarantee a transitively normal subgroup.