Direct factor is not finite-join-closed

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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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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.

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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.