Subnormality is not finite-join-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) not satisfying a subgroup metaproperty (i.e., finite-join-closed subgroup property).
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A join of two subnormal subgroups of a group need not be a subnormal subgroup. In fact, it might well be a proper contranormal subgroup: its normal closure might be the whole group.

Related facts

Partial truth of the opposite

A group satisfying subnormal join property is a group where a join of finitely many subnormal subgroups is subnormal. Every finite group, every Noetherian group, as well as every group satisfying ascending chain condition on subnormal subgroups satisfies the subnormal join property. For more, see ascending chain condition on subnormal subgroups implies subnormal join property.

A join-transitively subnormal subgroup is a subgroup whose join with any subnormal subgroup is again subnormal. There are many join-transitively subnormal subgroups. For instance:

There are other conditions under which joins of finitely many subnormal subgroups are subnormal:

Facts used

  1. Join of two 3-subnormal subgroups may be proper and contranormal


The proof follows directly from fact (1), or, more concretely, from the examples used to prove fact (1). Note that since 2-subnormal implies join-transitively subnormal, getting both subgroups to be 3-subnormal is the best we can achieve in terms of subnormal depth.