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I've been reading about Number Theory and I came across this proof that the finite subgroups of the multiplicative group of a field is cyclic. However, it seems the proof applies to all finite groups so please tell me where I go wrong.

Let $G$ be a generic group of order $n$. By Lagrange the order of the elements of the group must divide $n$. Let $\psi(d)$ count the number of elements in the group of order $d$. Since $\sum_{d|n} \psi(d)=n$ the Möbius Inversion theorem tells us that this function is just Euler's totient function. Since this function is always one or greater, we always have that there exists an element of order exactly $n$, so the group is cyclic.

Obviously I know the answer is no, but I thought using this title would make things more interesting.

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up vote 5 down vote accepted

What Möbius inversion tells you is that if a function $\psi$ satisfies $\sum_{d | n} \psi(d) = n$ for all $n$, then $\psi(d) = \varphi(d)$. But your function $\psi$ is not just a function of $d$, it is also a function $\psi_G$ of $G$, and as $G$ changes $\psi_G$ also changes. You used notation which hid the dependence on $G$, which made it harder to catch this mistake. (In general I am a fan of notation which makes various hidden dependencies like this explicit. My least favorite example of notation which does not do this is partial derivative notation, which tells you which variable to change but hides the dependence on which variables not to change.)

The proof that every finite subgroup of the multiplicative group of a field is cyclic also uses more than what you've written.

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Consider the direct sum of two copies of $\mathbb{Z}_2$.

The sum of the $\psi(d)$ is indeed $4$. There is no element of order $4$. The function $\psi$ is not the Euler $\varphi$-function. The formula from which you derived the "proof" by inversion does not hold at $2$.

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I mean I know the answer is false, but what is the error in the proof. – Steven-Owen Oct 9 '12 at 4:30
@ricky I think André's point is that since you know a simple example of a non-cyclic group, you can insert it into your proof and see exactly what goes wrong. – MJD Oct 9 '12 at 6:35
The error is that for Mobius inversion to work as you invoked it, it must be that $\sum_{d \mid m} \psi(d) = m$ for ALL $m$, but you've only shown it to hold for one particular $m$ (the order of the group). – Sridhar Ramesh Dec 3 '14 at 19:27

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