0
$\begingroup$

What are normal closure of a complement of a subgroup and intersection of all normal subgroups?

If X is a nonempty subset of a group G, define the normal closure X complement of X to be the intersection of all normal subgroups of G that contain X; that is, core H = {g in G | g in aHa^-1 for all a in G} = intersections of {aHa^-1 | a in G}

Sorry first time posting here and I'm not good with codes.

$\endgroup$
0
$\begingroup$

If $H$ is a proper subgroup of $G$, then the subgroup generated by its complement, $\langle G-H \rangle=G$ (this follows from the fact that a group cannot be the union of two proper subgroups). Hence the normal closure, $(G-H)^G=\langle G-H \rangle^G=G$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.