Nilpotent derived subgroup implies subnormal join property

Statement
Any group whose derived subgroup is a nilpotent group satisfies the subnormal join property: a join of subnormal subgroups of the group is subnormal.

Facts used

 * 1) uses::Nilpotent implies every subgroup is subnormal
 * 2) uses::Join of subnormal subgroups is subnormal iff their commutator is subnormal

Proof
Given: A group $$G$$ such that $$G' = [G,G]$$ is nilpotent. Subnormal subgroups $$H,K$$ of $$G$$.

To prove: $$\langle H, K \rangle$$ is subnormal.

Proof: Clearly, $$[H,K] \le [G,G]$$. Since $$[G,G]$$ is nilpotent by assumption, fact (1) tells us that $$[H,K]$$ is a subnormal subgroup of $$[G,G]$$. Further, $$[G,G]$$ is a normal subgroup of $$G$$, so $$[H,K]$$ is subnormal in $$G$$. Thus, by fact (2), $$\langle H, K \rangle$$ issubnormal in $$G$$.

Textbook references

 * , Page 388, Theorem 13.1.7, Section 13.1 (Joins and intersections of subnormal subgroups)