Central factor is upper join-closed

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

Statement with symbols

Suppose G is a group, H is a subgroup of G, and K_i, i \in I, are subgroups of G all containing H. Suppose, further, that H is a central factor of each K_i. Then, H is also a central factor of the join of the K_is.

Related facts

Proof

Using function restriction expressions

This subgroup property implication can be proved by using function restriction expressions for the subgroup properties
View other implications proved this way |read a survey article on the topic

Given: G is a group, H is a subgroup of G, and K_i, i \in I, are subgroups of G all containing H. H is a central factor of each K_i.

To prove: H is a central factor of the join of K_is. In other words, if g is an element in the join of the K_is, then there exists h \in H such that conjugation by g agrees with conjugation by h on H.

Proof: Suppose g is an element in the join of the K_is. Then, there exist i_1, i_2, i_3, \dots, i_n and g_1 \in K_{i_1}, g_2 \in K_{i_2}, \dots, g_n \in K_{i_n} such that g = g_1g_2 \dots g_n.

For each g_j, since H is a central factor of K_{i_j}, there exists h_j \in H such that conjugation by h_j agrees with conjugation by g_j on H. Set h = h_1h_2 \dots h_n. Then, since conjugation is a group action, we see that conjugation by h agrees with conjugation by g on H, completing the proof.

Proof using centralizer definition

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