Central factor is upper join-closed

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 $K_{i}$s.

Related facts

• Central factor satisfies intermediate subgroup condition
• Central factor satisfies image condition
• Central factor is transitive
• Central factor does not satisfy transfer condition

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 $K_{i}$s. In other words, if $g$ is an element in the join of the $K_{i}$s, 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 $K_{i}$s. 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 $g=g_{1}g_{2}\dots g_{n}$.

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 $h=h_{1}h_{2}\dots h_{n}$. 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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