# 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.

## 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.