Derived subgroup satisfies ascending chain condition on subnormal subgroups implies subnormal join property

Statement
Suppose $$G$$ is a group with the property that the fact about::derived subgroup (i.e., commutator subgroup) $$[G,G] = G'$$ satisfies the ascending chain condition on subnormal subgroups. Then, if $$H$$ and $$K$$ are fact about::subnormal subgroups of $$G$$, so is the join $$\langle H, K \rangle$$.

Facts used

 * 1) uses::Normality is commutator-closed
 * 2) uses::Normality satisfies intermediate subgroup condition
 * 3) uses::Ascending chain condition on subnormal subgroups is normal subgroup-closed
 * 4) uses::Join-transitively subnormal of normal implies finite-conjugate-join-closed subnormal
 * 5) uses::Product with commutator equals join with conjugate: This states two things: if $$H \le G$$ and $$g \in G$$, then $$H$$ normalizes $$[g,H]$$, and further, $$H[g,H] = \langle H, H^g \rangle$$.
 * 6) uses::Join of subnormal subgroups is subnormal iff their commutator is subnormal

Proof of subnormal join property
We prove the statement by induction on the subnormal depth of one of the two subgroups whose join we are taking. Specifically, we prove the following statement by induction on $$h$$, for $$h$$ a nonnegative integer.

Formulation to be proved by induction on $$h \ge 0$$: Any subgroup of subnormal depth $$h$$ in a group whose derived subgroup satisfies the ascending chain condition on subnormal subgroups is join-transitively subnormal. In long form: Suppose $$G$$ is a group whose derived subgroup $$[G,G]$$ satisfies the ascending chain condition on subnormal subgroups, $$H$$ is a subnormal subgroup of $$G$$ of subnormal depth $$h$$, and $$K$$ is a subnormal subgroup of $$G$$ of subnormal depth $$k$$. Then, the join of subgroups $$\langle H, K \rangle$$ is also a subnormal subgroup of $$G$$.

Note that when we apply this induction, the ambient group for which we use the inductive hypothesis is not the same as the ambient group for which we want to prove the goal of the inductive step.

Base case for induction: This is the case $$h = 0$$. In this case, $$H = G$$, so $$\langle H, K \rangle = G$$ is also subnormal. This settles the base case.

Inductive step ($$h - 1 \to h$$):

Inductive hypothesis: In any group whose derived subgroup satisfies ascending chain condition on subnormal subgroups, any subnormal subgroup of subnormal depth at most $$h - 1$$ satisfies the ascending chain condition on subnormal subgroups. In other words, the join of any two subnormal subgroups is subnormal if either of them has subnormal depth at most $$h - 1$$.

Inductive goal:

Given: $$G$$ is a group such that the derived subgroup $$[G,G]$$ satisfying the ascending chain condition on subnormal subgroups. $$H$$ is a subgroup of subnormal depth $$h$$ in $$G$$ and $$K$$ is a subnormal subgroup of $$G$$.

To prove: $$\langle H, K \rangle$$ is a subnormal subgroup of $$G$$.

Proof: The key proof idea is to use the inductive step within a smaller ambient group, namely the normal closure of the subgroup we are starting out with, and then bootstrap from this to the bigger group. There are two key places where the ascending chain condition on subnormal subgroups is used. The first is that this property is a normal subgroup-closed group property (Fact (3)), allowing the shift down to the normal closure. The second is that the property allows us to go from commutators with finite subsets to commutators with an entire group.

Notice that the proof that $$[H,K]$$ is subnormal depends only on the fact that $$H$$ is subnormal, and does not require the subnormality of $$K$$ -- it is only the last step where we use subnormality of $$K$$.

Textbook references

 * , Page 388, Theorem 13.1.8, Section 13.1 (Joins and intersections of subnormal subgroups)