# 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.

## 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.