View source for Direct factor is not finite-join-closed

{{subgroup metaproperty dissatisfaction|
property = direct factor|
metaproperty = finite-join-closed subgroup property|
first corollary = join-closed subgroup property|
second corollary = strongly finite-join-closed subgroup property|
third corollary = strongly join-closed subgroup property}}

==Statement==

A join of finitely many [[direct factor]]s of a group need not be a direct factor. More specifically, it is possible to have a [[group]] <math>G</math> and two subgroups <math>H, K</math> of <math>G</math> such that both <math>H</math> and <math>K</math> are direct factors and the join <math>HK</math> 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 <math>C_n</math> denotes the cyclic group of order <math>n</math>. Define:

<math>G = C_4 \times C_2</math>.

Consider the following subgroups:

<math>H = 0 \times C_2, \qquad K = \{ (2,1), (0,0) \}, L = C_4 \times 0</math>.

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

<math>HK = \{ (2,1), (2,0), (0,1), (0,0) \}</math>.

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