Normality is upper join-closed

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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 \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_is.

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

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_is.

Proof: Let K be the join of the K_is. 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.