Originally Posted by

**slevvio** Hello all. I have something in which there must be a contradiction but I cannot find it.

So let $\displaystyle \phi $ be the usual Euler phi/totient function, and let $\displaystyle G$ be a finite group. Then we have

(1) $\displaystyle \phi_G(d) = \phi(d) \cdot q(d)$

where $\displaystyle q(d)$ = the number of cyclic subgroups of order $\displaystyle d$ and $\displaystyle \phi_G(d)$ = the number of elements of order $\displaystyle d$ in $\displaystyle G$.

Then we also have the theorem

(2) $\displaystyle \displaystyle\sum_{d | n} \phi(d) = n.$

But I have a theorem in my notes

which states $\displaystyle \displaystyle\sum_{d | |G|} \phi_G (d) = |G|$. But by (1), this is equivalent to the statement $\displaystyle \displaystyle\sum_{d | |G|} \phi(d) q(d) = |G|$.

But doesn't this contradict (2) ? Unless $\displaystyle q(d) = 1 \text{ whenever } \phi(d) \not= 0$ which cannot be true otherwise there would be no point looking at $\displaystyle \phi_G(d)$. Can anyone help?