Mobius Inversion for any Arithmetic Function My question is whether or not it is valid, (or more importantly, is there any point in doing so?) defining a function $g(n)$ as follows, and if it is, does this mean that all arithmetic functions must have equivalent expressions to one another that involves the Mobius function? 
$$\delta \left( x,y \right) =\cases{1&$x=y$\cr 0&$x\neq y$\cr}$$
$$g \left( n \right) =\sum _{k=1}^{n}f \left( k \right) \delta \left( {
\frac {n}{k}}, \Bigl\lfloor {\frac {n}{k}} \Bigr\rfloor  \right) 
$$
$$f  \left( n \right) =\sum _{k=1}^{n}\mu \left( k \right) g \left( {
\frac {n}{k}} \right) \delta \left( {\frac {n}{k}}, \Bigl\lfloor
{\frac {n}{k}} \Bigr\rfloor \right) 
$$
I also don't understand how mobius inversion applies to any abelian group, and want to know the proof for this. And the proof of mobius inversion on wikipedia here I am also unclear about, in particular how the author moves from the 3rd to forth line, eliminating the indicator function to arrive at the arithmetic sum of the Mobius function.
 A: I’m surprised that no one else has answered this before me.
First, your notation $$\delta\left(\frac nk,\left\lfloor\frac nk\right\rfloor \right)$$ simply gives $1$ when $k|n$, $0$ otherwise. Therefore your notation $$\sum_{k=1}^nf(k)\delta\left(\frac nk,\left\lfloor\frac nk\right\rfloor \right)$$
is much more economically written $\sum_{d|n}f(d)$, and the third line simply restates the Möbius Inversion Formula.
I think that the Formula is best understood when translated into the language of arithmetic functions $f:\Bbb N\to\Bbb Z$, (here $\Bbb N$ is the positive integers) where as usual, addition is done pointwise but the “multiplication” is “$*$”, defined by $$(f*g)(n)=\sum_{d|n}f(d)g(n/d)\,.$$
Then, it’s necessary to show associativity of this multiplication, but this poses no problem. There is a neutral element (identity) for $*$, namely $\Bbb I(n)=\delta(1,n)$, the function that’s zero everywhere but at $n=1$. If we define $Z(n)=1$ for all $n\ge1$, then this is an arithmetic function too, and Möbius Inversion says just that if $g=Z*f$, then $f=\mu*g$. Stare at this, and it says that $Z*\mu=\Bbb I$, in other words $\mu$ and $Z$ are inverses of each other.

How now to interpret Möbius to apply to functions from $\Bbb N$ to any abelian group ($\Bbb Z$-module) $M$? Given our module $M$, we can form the set of “$M$-etic functions”, all $F:\Bbb N\to M$, call the set $M^{\Bbb N}$, on which $Z$ acts in the usual way, but also is a module over the set of arithmetic functions (which we can now call $\Bbb Z^{\Bbb N}$). The operation is by the same convolution formula, if $f\in\Bbb Z^{\Bbb N}$ and $F\in M^{\Bbb N}$, then $f*F(n)=\sum_{d|n}f(d)F(n/d)$. You have to verify that $f*(g*F)=(f*g)*F$ and $\Bbb I*F=F$ and $f*(F+G)=f*F+f*G$, and also I guess that $(f+g)*F=f*F+g*F$.
Finally, with this notation out of the way, you get from the hypothesis $G(n)=\sum_{d|n}F(d)$, rewordable as $G=Z*F$, to the conclusion $\mu*G=F$ just by multiplying both sides on the left by the inverse of $Z$, namely $\mu$.
Here’s a nice application, where $M$ is the multiplicative group of nonzero rational functions in a variable $X$, over $\Bbb Z$. Let $n>0$. Then the polynomial $X^n-1$ has for its roots all the $n$-th roots of unity, some of which are primitive $d$-th roots for $d|n$. If we call $\Phi_d$ the polynomial whose roots are the primitive $d$-th roots of unity, we get $\Phi_d(X)\in\Bbb Z[X]$; it’s the $d$-th cyclotomic polynomial. And we have $$X^n-1=\prod_{d|n}\Phi_d(X)\,,$$
and from Möbius Inversion, we get $$\Phi_n(X)=\prod_{d|n}(X^d-1)^{\mu(n/d)}\,.$$
