Nilpotent derived subgroup implies subnormal join property

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group with nilpotent derived subgroup) must also satisfy the second group property (i.e., group satisfying subnormal join property)
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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. Nilpotent implies every subgroup is subnormal
2. 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, $[H,K]\leq [G,G]$. 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 $G$.

References

Textbook references

• A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, ^{More info}, Page 388, Theorem 13.1.7, Section 13.1 (Joins and intersections of subnormal subgroups)