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This question was originally posted in Elements of finite order in the group of arithmetic functions under Dirichlet convolution.

and it goes as follows: Let G be the group consisting of all arithmetic functions (i.e. functions $f:\mathbb{N} \to \mathbb{C}$) under the "convolution"operation $\ast,$ defined as $$(f\ast g)(n):=\sum_{ab=n}f(a)g(b), n \in \mathbb{N}.$$ (Note that each function in the group assigns the value $1$ to $1.$)

Alexander Gruber showed in the aforementioned post that G is a torsion-free abelian group. A few further facts can be easily proved:

(1) G is not finitely generated and in particular is infinite.

(2)If one lets H to denote the subgroup of G consisted of multiplicative functions, then $|G:H|=\infty.$

However my question is: What is known in the literature about the group structure of G ?

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$f(1) = 1$ isn't part of the definition of an arithmetic function; please include this condition explicitly in the definition of $G$ (or else it isn't a group). – Qiaochu Yuan Dec 29 '12 at 7:08

$G$ is a certain subgroup of the multiplicative group of the ring of all arithmetic functions under convolution and pointwise addition. This ring is isomorphic to the ring of formal Dirichlet series, which is in turn isomorphic to the ring of formal power series $\mathbb{C}[[x_1, x_2, ... ]]$ in countably many variables, one for each prime (write out $\frac{1}{n^s}$ in terms of the prime factorization of $n$).

So the question reduces to the following: what is the structure of the group of complex power series in countably many variables with constant term $1$ under multiplication?

To answer that question, observe that the logarithm of such a formal power series is well-defined using the formal power series of $\log (1 + x)$, and the result is a formal power series with constant term $0$, and conversely. Hence the group in question is isomorphic to the additive group of formal power series in countably many variables with constant term $0$, which is just a countable product of copies of $\mathbb{C}$, or in purely group-theoretic terms a vector space over $\mathbb{Q}$ of dimension the cardinality of $\mathbb{R}$.

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"purely group-theoretic terms"? – the_fox Dec 29 '12 at 8:50
@Stefanos: "countable product of copies of $\mathbb{C}$" describes a $\mathbb{C}$-vector space, but the OP is only interested in the group structure. (Maybe you're confused about "vector space over $\mathbb{Q}$"? This is a property, not a structure.) – Qiaochu Yuan Dec 29 '12 at 9:20
I don't think I am confused. I'm only observing that "pure group theory" and "vector spaces" don't mix. – the_fox Jan 3 '13 at 2:33
@Stefanos: sure they do. First of all, vector spaces are groups. Second of all, "vector space over $\mathbb{Q}$" is a group-theoretic condition on an abelian group. It means that the group is not only divisible but that it is a coproduct of copies of $\mathbb{Q}$. In particular, specifying the action of $\mathbb{Q}$ is not additional data; if such an action exists, it is unique. – Qiaochu Yuan Jan 3 '13 at 2:42
I am sorry, but vector spaces are not groups. – the_fox Jan 3 '13 at 3:14

Your definition must include that $f(1)\not= 0$, because $f^{-1}(n)$ is recursively defined in terms of $1/f(1)$. (The formula can be found on the Wikipedia.) So, denote by $\mathcal{A}$ the group of all arithmetic functions with $f(1)\not= 0$.

Let $\mathcal{U}$ be the subgroup of all $f\in \mathcal{A}$ such that $f(1)=1$ and $\mathcal{C}$ be the subgroup of scalar functions $\epsilon_c(n):=c\epsilon(n)$, where $\epsilon=\text{id}_\mathcal{A}$. Note that $\mathcal{C}\cap \mathcal{U}=\{\epsilon\}$ Define $(\frac{1}{c}f)(n):=\frac{f(n)}{c}$. If we consider an arbitrary $g\in \mathcal{A}$, then $\frac{1}{f(1)}f\in\mathcal{U}$. We have $$\left(\frac{1}{g(1)}g\star \epsilon_{g(1)}\right)(n)=\sum_{ab=n}\frac{1}{f(1)}f(a)f(1)\epsilon(b)=(f\star \epsilon)(n)=f(n).$$ Thus we can uniquely factor $f=\frac{1}{f(1)}f\star \epsilon_f(1)$, whence $\mathcal{A}=\mathcal{U}\oplus \mathcal{C}$.

Now define $\mathcal{K}\subseteq \mathcal{U}$ to be the subset of (not necessarily multiplicative) functions which vanish on prime powers. Since $f^{-1}(n)$ is defined in terms of $f(n)$, and since $\mathcal{K}$ is obviously closed under $\star$, it follows immediately that $\mathcal{K}$ is a subgroup of $\mathcal{U}$. Again $\mathcal{H}\cap \mathcal{K}=\{\epsilon\}$, where $\mathcal{H}$, as given by OP, is the subgroup of multiplicative functions. For $f\in \mathcal{U}$, define $\Pi_f$ so that $$\Pi_f(p_1^{e_1}\cdots p_s^{e_s})=\prod_{k=1}^r f(p_k^{e_k}).$$ Obviously $\Pi_f\in \mathcal{H}$. We have that $$\begin{eqnarray*} \left(\Pi_f^{-1}\star f \right)(p^{e})&=&\sum_{k=0}^e\Pi_f(p^k)^{-1}f(p^{e-k})\\&=&\sum_{k=0}^e\Pi_f^{-1}(p^k)\Pi_f(p^{e-k})\\&=&(\Pi_f^{-1}\star\Pi_f)(p^e)\\&=&\epsilon(p^e)\\&=&0.\end{eqnarray*}$$ It follows that $\Pi_f^{-1}\star f\in \mathcal{K}$. Now of course we can write $f=\Pi_f\star\left(\Pi_f^{-1}\star f\right)$, so $f$ uniquely factors into $\mathcal{H}$ and $\mathcal{K}$ parts, and we have that $\mathcal{U}=\mathcal{H}\oplus \mathcal{K}$.

So we have obtained that $\mathcal{A}=\mathcal{H}\oplus \mathcal{K}\oplus \mathcal{C}$.

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