Join of finitely many direct factors implies central factor

Statement with symbols
Suppose $$G$$ is a group and $$H_1, H_2, \dots, H_n$$ $$n \ge 0$$) are direct factors of $$G$$. Then, the join of the $$H_i$$s, which is also the product $$H_1H_2\dots H_n$$ (because direct factor implies normal) is a central factor of $$G$$.

Facts used

 * 1) uses::Trivial subgroup is central factor
 * 2) uses::Direct factor implies join-transitively central factor: The join of a direct factor and a central factor is a central factor.

Proof
Given: $$G$$ is a group and $$H_1, H_2, \dots, H_n$$ are direct factors of $$G$$.

To prove: The join of all the $$H_i$$s is a central factor of $$G$$.

Proof: We prove this by induction on $$n$$, starting with the case $$n = 0$$. In the case $$n = 0$$, the join is the trivial subgroup, so fact (1) completes the proof.

Suppose the statement is true for $$n - 1$$, and we want to prove it for $$n$$. Since the statement is true for $$n - 1$$, we know that the join of $$H_1, H_2, \dots, H_{n-1}$$ is a central factor. By fact (2), the join of this with $$H_n$$ must also be a central factor. But this latter join is the same as the join of $$H_1, H_2, \dots, H_n$$. This completes the proof.