Subnormality is not intersection-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) not satisfying a subgroup metaproperty (i.e., intersection-closed subgroup property).
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Statement

An arbitrary intersection of subnormal subgroups of a group need not be subnormal.

Related facts

• Subnormality is finite-intersection-closed
• Normality is strongly intersection-closed
• 2-subnormality is strongly intersection-closed
• Subnormality of fixed depth is strongly intersection-closed

Facts used

1. Descendant not implies subnormal

Proof

By the definition of descendant subgroup, it is clear that if an intersection of subnormal subgroups were subnormal, then a descendant subgroup would always be subnormal. Thus, fact (1) shows that an intersection of subnormal subgroups need not be subnormal.

Specifically, we can take a group with a descendant subgroup that is not subnormal, and look at a descendant series for it. The first non-subnormal member of this series arises as the intersection of the previous members, which are subnormal, thus yielding an intersection of subnormal subgroups that is not subnormal.