# Ascending chain condition on subnormal subgroups 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 satisfying ascending chain condition on subnormal subgroups) must also satisfy the second group property (i.e., group satisfying subnormal join property)
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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 satisfying ascending chain condition on subnormal subgroups) must also satisfy the second group property (i.e., group satisfying generalized subnormal join property)
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Get more facts about group satisfying ascending chain condition on subnormal subgroups|Get more facts about group satisfying generalized subnormal join property

## Statement

Any group satisfying ascending chain condition on subnormal subgroups, i.e., any group in which there is no infinite ascending chain of subnormal subgroups, also satisfies the following two conditions:

1. It is a group satisfying subnormal join property, i.e., the join of any two (and hence finitely many) subnormal subgroups of the group is again subnormal.
2. It is a group satisfying generalized subnormal join property, i.e., the join of any, possibly infinite, collection of subnormal subgroups of the group is again subnormal.

## Proof

### Proof of subnormal join property

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This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).

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 satisfying the ascending chain condition on subnormal subgroups is join-transitively subnormal. In long form: Suppose $G$ 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 a group satisfying the ascending chain condition on subnormal subgroups, any subnormal subgroup of subnormal depth at most $h - 1$ is join-transitively subnormal. In other words, in any group satisfying ascending chain condition on subnormal subgroups, 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 satisfying the ascending chain condition on subnormal subgroups. $H$ is a subgroup of subnormal depth $h$ in $G$

To prove: $H$ is join-transitively subnormal in $G$.

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 (1)), allowing the shift down to the normal closure. The second is that the property allows us to go from joins of finitely many conjugate subgroups to joins of infinitely many conjugate subgroups (Fact (3)).

Step no. Assertion/construction Facts used Given data/assumptions used Previous steps used Explanation Commentary
1 Let $L = H^G$ be the normal closure of $H$ in $G$. In particular, $L$ is normal and is generated by all the conjugate subgroups of $H$. $H \le G$ Preparing ground for shifting ambient group down to normal closure.
2 $H$ is $(h - 1)$-subnormal in $L$ definition of subnormal depth $H$ is $h$-subnormal in $G$. Step (1) [SHOW MORE] Fixing subnormality in $L$.
3 $L$ is a group satisfying ascending chain condition on subnormal subgroups Fact (1) $G$ satisfies ascending chain condition on subnormal subgroups Step (1) Fact+Step+Given direct Fixing ambient group property of $L$.
5 $H$ is join-transitively subnormal in $L$, i.e., the join of $H$ and any subnormal subgroup of $L$ is subnormal in $L$. inductive hypothesis Steps (2), (3) [SHOW MORE] Apply inductive hypothesis.
6 $H$ is finite-conjugate-join-closed subnormal in $G$, i.e., a join of finitely many conjugates of $H$ in $G$ is subnormal in $G$. Fact (2) Step (5) Fact+Step direct. Move to ambient group $G$.
7 $H$ is conjugate-join-closed subnormal in $G$, i.e., a join of arbitrarily many conjugates of $H$ in $G$ is subnormal in $G$. Fact (3) $G$ satisfies ascending chain condition on subnormal subgroups. Step (6) [SHOW MORE] Key step going from finite joins to arbitrary joins, use ascending chain condition on subnormal subgroups.
8 $H$ is join-transitively subnormal in $G$, i.e., a join of $H$ and any subnormal subgroup of $G$ is subnormal in $G$. Fact (4) Step (7) Fact+Step direct Clinch.

### Proof of generalized subnormal join property

This follows directly from the ascending chain condition and the subnormal join property.