Normality is upper join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) satisfying a subgroup metaproperty (i.e., upper join-closed subgroup property)
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Statement

Statement with symbols

Suppose $H$ is a subgroup of $G$ , $I$ is a nonempty indexing set, and $K_{i}, i \in I$ are subgroups of $G$ containing $H$ , such that $H ◃ K_{i}$ (i.e., $H$ is a normal subgroup of $K_{i}$ ) for each $i \in I$ . Then, $H$ is normal in the join of the $K_{i}$ s.

Related facts

Related facts about normality

Related facts about upper join-closedness

The fact about normality generalizes to the following:

Left-inner right-monoidal implies upper join-closed: A subgroup property that has a function restriction expression with the left property being inner automorphisms and the right property being monoidal (closed under composition) is upper join-closed.

Other manifestations of the general fact include:

Here are some related properties that are not upper join-closed:

Analogues and breakdowns of analogues in other algebraic structures

Ideal property is upper join-closed for Lie rings: If $I$ is a subring of a Lie ring $L$ such that $I$ is an ideal in two subrings $A_{j} \leq L$ , where $j \in J$ , an indexing set, then $I$ is also an ideal in the Lie subring generated by all the $A_{j}$ s.
Ideal property is not upper join-closed for alternating rings
Normality is not upper join-closed for algebra loops

Proof

Given: A group $G$ , a subgroup $H$ , a nonempty indexing set $I$ , and a collection of subgroups $K_{i}, i \in I$ , such that $H$ is normal in $K_{i}$ for each $i \in I$ .

To prove: $H$ is normal in the join of the $K_{i}$ s.

Proof: Let $K$ be the join of the $K_{i}$ s. For $g \in K$ , we can write:

$g = g_{1} g_{2} g_{3} \dots g_{n}$

where $g_{j} \in K_{i_{j}}$ for some index element $i_{j}$ . Thus, if $c_{g}$ denotes conjugation by $g$ , we have:

$c_{g} = c_{g_{1}} \circ c_{g_{2}} \circ \dots \circ c_{g_{n}}$

Now, since $H$ is normal in $K_{i_{j}}$ , each $c_{g_{j}}$ acts as an automorphism of $H$ . Thus, their composite, namely $c_{g}$ , is also an automorphism of $H$ . In other words, $c_{g} (H) = H$ for every $g \in K$ , showing that $H$ is normal in $K$ .