Direct factor is not finite-join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., direct factor) not satisfying a subgroup metaproperty (i.e., finite-join-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Join-closed subgroup property (?), .
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Statement

A join of finitely many direct factors of a group need not be a direct factor. More specifically, it is possible to have a group $G$ and two subgroups $H,K$ of $G$ such that both $H$ and $K$ are direct factors and the join $HK$ is not a direct factor.

Related facts

• Transitive normality is not finite-join-closed
• Central factor is not finite-join-closed

Proof

An abelian group example

Suppose $C_{n}$ denotes the cyclic group of order $n$. Define:

$G=C_{4}\times C_{2}$.

Consider the following subgroups:

$H=0\times C_{2},\qquad K=\{(2,1),(0,0)\},L=C_{4}\times 0$.

Then, both $H$ and $K$ are direct factors of $G$, with a common direct factor complement $L$. On the other hand, we have:

$HK=\{(2,1),(2,0),(0,1),(0,0)\}$.

This is not a direct factor of $G$, because if a complement exists, it must have order two, but all elements of $G$ outside $HK$ have order four.