Actually, it turns out that may be 0 for some .
For example: a finite group is not cyclic iff .
Hello all. I have something in which there must be a contradiction but I cannot find it.
So let be the usual Euler phi/totient function, and let be a finite group. Then we have
(1)
where = the number of cyclic subgroups of order and = the number of elements of order in .
Then we also have the theorem
(2)
But I have a theorem in my notes
which states . But by (1), this is equivalent to the statement .
But doesn't this contradict (2) ? Unless which cannot be true otherwise there would be no point looking at . Can anyone help?
I don't particularly understand you're confusion, I don't see an immediate contradiction. One way that someone could define this is to define the relation on by if and only if . If one then calls the equivalence classes of this (clearly equivalence) relation then one has that where is a set which contains one representative from each equivalence class. It follows then that but it's not hard to prove that (using your notation) and thus, we may conclude that but it shouldn't be hard to convince yourself that this can be written (recalling Lagrange's theorem and the fact that even if the group does not contain any cyclic subgroups of order then one may add the term since it's zero) as . So, this in particular proves that by considering that has exactly one subgroup (and of course this implies one cyclic subgroup) for each divisor of . How does using these facts contradict anything?