Subnormality is not intersection-closed

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) uses::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.