Central factor is upper join-closed

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
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.