Subgroup with order =LCM of two subgroups The following is a question that was asked by my teacher as a ponder-upon question:to which unfortunately I have not been able to put a single forward step.
If an abelian group has subgroups of order $m\ \&\ n$, then is it true that it must have a subgroup of order $= LCM(m,n)$?
I am a beginner in group theory, so please consider explaining in detail your answer. Any reference to the proof is okay as well, but please don't down vote if its too simple.

For abelian groups converse of langranges theorm holds.Hence corrresponding to each divisor of of order of the group there exsists a subgroup.
