When it comes to studying normal subgroups, how would one go about proving the following?
Prove that if $N\triangleleft G$ and $G/N$ is cyclic, then there exists a subgroup $N'\triangleleft G$ such that the index $[G:N']=p$, where $p$ is a prime.
Is there a straightforward way of going about this? Thanks for any suggestions!
It follows from the fact that the fourth isomorphism theorem links normal subgroups of $ G/N $ to normal subgroups of $ G $ containing $ N $. As $ G/N $ is cyclic, there is a subgroup $ N'/N $ of $ G/N $ (which is normal, since $ G/N $ is cyclic, thus abelian) of index $ p $ where $ p $ is a prime dividing the order of $ G/N $; and the subgroup $ N' $ of $ G $ is normal and has index $ p $ in $ G $. (This only works if $ G/N $ is nontrivial, i.e if $ G \neq N $ - otherwise, the claim is clearly false.)