I'm very new to ACL, and at the moment I've been trying to set up a shared folder such that any user belonging to a certain group (lets call it "thegroup") can have full access to the folder (which includes read, write, execute, and "creation" of files). When a user belonging to a group creates a file, the ownership should be automatically set to the user as well as "thegroup", and the file should be accessible in every single way by other members of the "thegroup". I'm not sure what the user ownership of the folder should be called so I just decided to set it to root.   I've been following the instructions here but so far I can only list the files located in the directory as a normal user. I do not have read, write, or creation capabilities after doing the following commands:  mkdir folder setfacl -m u:root:7 folder setfacl -m -d u:root:7 folder setfacl -m g:thegroup:7 folder setfacl -m -d g:thegroup:7 folder   From what I understood, by setting the default ownership all the files that are created inside the folder should inherit the parent folder's permission, right? However, this doesn't seem to be the case.  Is there something I'm missing?  Note: the debian manual mentions that for each user that I want to add access to I have to explicitly add him to the ACL. I'm trying to avoid this by simply adding the user to a group, and then giving the group full access to the folder. Is this possible with ACL, or do I have no choice but to manually specify every user that's going to access the folder?
I want to set a folder such that anything created within it (directories, files) inherit default permissions and group.    Lets call the group "media".  And also, the folders/files created within the directory should have g+rw automatically.
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Changing isometric drawing's constraint angles in Illustrator Let $G$ be a finite group, say a group with $n$ elements, and let $\emptyset \neq S \subseteq G$. Suppose $e \in S$, and that $S$ is closed with respect to multiplication. Prove that $S$ is a subgroup of $G$.    Let $G = \{a_1, \ldots, a_n\}.$ Let $a_i \in G.$ Then any number of $a_ia_1, a_ia_2, \ldots, a_ia_n \in S,$ since $S$ is closed under multiplication. One of those $a_ia_j$ is $e$ since $e \in S.$ Then either $a_i$ or $a_j$ is $a^{-1}.$ Would this argument work?