Normality is upper join-closed

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 $$H \triangleleft K_i$$ (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 about normality

 * Join lemma for normal subgroup of subgroup with normal subgroup of whole group
 * Normality is not UL-join-closed
 * Normality is strongly join-closed
 * Normality is strongly intersection-closed
 * Normality is UL-intersection-closed
 * Normality satisfies intermediate subgroup condition
 * Normality satisfies transfer condition

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:


 * Transitive normality is upper join-closed
 * Central factor is upper join-closed

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


 * Characteristicity is not upper join-closed
 * Conjugacy-closedness is not upper join-closed
 * Subnormality is not finite-upper join-closed, subnormality is not permuting upper join-closed
 * 2-subnormality is not finite-upper join-closed, 2-subnormality is not permuting 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 $$A_j\le L$$, 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:

$$g = g_1g_2g_3\dots g_n$$

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 $$K$$.