If a set $S$ is not closed under some binary operation $\star$, is it true that $S$ cannot be a group under $\star$?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Did you figure out your dihedral group question? I was writing an answer to it. $\endgroup$– anonCommented Oct 9, 2013 at 14:36
-
$\begingroup$ Sorry, yeah I figured it out after staring at $D_8$'s lattice structure for a bit. $\endgroup$– AlexCommented Oct 9, 2013 at 14:38
-
1$\begingroup$ If you're looking at the lattice structure of $D_8$ hosted on a website in order to figure out $D_{2n}/\langle r\rangle$, it may be problematic; it's necessary to be able to figure out what quotient groups are and what their elements look like on your own, especially for basic groups like dihedral ones. Here is what I was going to write. $\endgroup$– anonCommented Oct 9, 2013 at 15:12
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
It is true. An operation in a group has to stay within the group.