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Let $(G, ∗)$ be the group of arithmetic functions $f : N \to C$ that satisfy $f (1)\neq 0$, with group operation given by the Dirichlet product $∗$. The identity function $I$ is the identity element of $G$ defined by $I : N \to Z$ where $I=1$ if $n=1$, $0$ if $n>1$.

An element $f \in G$ has finite order if there exists an $n\in N$ such that $f^n =I$ (here $f^n =f∗f∗\cdots ∗f$ with $n$ factors). Find all number theoretic functions $f : N \to Z$ with $f(1)\neq 0$ that have finite order in the group $G$.

Well now, I thought to start from the fact that since $f(1)\neq 0$, then $f$ has an inverse called $g$. I know that $f*g=I$ under Dirichlet product so if I need to find $f$ such that $f*f*f\cdots *f =I$ means that $f^{(n-1)}$ is my inverse function but that's just what I can came up with after hours, with the info I have from my lecture notes . If anyone can hep would be much appreciated.

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Alexander's proof shows that it is an abelian, non-torsion group. Letting H be the subgroup of multiplicative functions, it is easy to prove by contradiction that 1. H (and hence G) has infinite rank. 2. The group G/H has infinite rank. My question is : Is there a reference for the group structure of G or G/H ? –  Captain Darling Dec 28 '12 at 20:43
    
@Captain: you should ask this as a separate question. –  Qiaochu Yuan Dec 29 '12 at 3:12

2 Answers 2

up vote 2 down vote accepted

For the benefit of other readers, a number theoretic function is a function whose domain is the set of positive integers, and the Dirichlet product is $$(f\star g)(n)=\sum_{d\mid n}f\left(d\right)g\left(\frac{n}{d}\right)=\sum_{ab= n}f(a)f(b).$$

First, let me prove a couple lemmas.

Claim. The identity in $\langle G,\star\rangle$ is $$e(n):=\left\{\begin{array}{lcl}1&:&n=1\\ 0&:&n> 1\end{array}\right.$$ Proof.

Since $e$ vanishes whenever $d\not= n$, $(f\star e)(n)=\sum_{d\mid n} f\left(d\right)e\left(\frac{n}{d}\right)=f(n)$.

Claim. $$f^m(n)=\sum_{a_1a_2\ldots a_m=n} f(a_1)f(a_2)\cdots f(a_m).$$ Proof.

By induction, $$\begin{eqnarray*}f^m(n)=(f\star f^{m-1})&=&\sum_{a_1b=n}f(a_1)f^{m-1}(b)\\&=&\sum_{a_1b=n}f(a_1)\left(\sum_{a_2\cdots a_{m}=b}f(a_2)\cdots f(a_m)\right)\\&=&\sum_{a_1a_2\cdots a_m=n} f(a_1)f(a_2)\ldots f(a_m)\end{eqnarray*}.$$

Alright, let's get down to business. Let $f$ be an element of $\langle G, \star \rangle$ with finite order $m$.

Since $f^m=e$, $f^m(1)$ needs to be $1$. By the second claim, $f^m(1)=f(1)^m$, so it follows that $f(1)=1$. Next, $f^m(n)$ needs to be $0$ for any $n\not= 1$. Take the smallest $n>1$ for which $f(n)\not= 0$. We have that $$f^m(n)=mf(1)^{m-1}f(n)+ \sum_{a_1\ldots a_m=n}_{a_1,\ldots,a_m<n} f(a_1)\ldots f(a_m)=mf(n)=0.$$ We assumed that $f(n)\not=0$, so we must have that $m=0$, but contradicts that $f$ has finite order. So $f(n)=0$ for all $n> 1$.

Thus we have that the only $f$ with finite order $e$, so $\langle G, \star \rangle$ is torsion-free.

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If you know the characterization of the product via formal Dirichlet series, I'm pretty sure you can also note that $$F(s)=\sum_n {f(n)\over n^s}$$ gives $$\underbrace{f*f*f*\ldots *f}_{m\text{ times}}=e$$ iff $F^m(s)\equiv 1$, where the power indicates normal multiplication and the $*$ DENOTES convolution.

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