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)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about central factor |Get facts that use property satisfaction of central factor | Get facts that use property satisfaction of central factor|Get more facts about upper join-closed subgroup property

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 $g = g_1g_2 \dots g_n$.

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

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.