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Let $(G,*)$ be the group of number theoretic functions $f$ with $f(1)\not =0$.

1)Show that if $f$ is a multiplicative function and $f$ is not identically zero, then $f\in G$.

2) Show that the Dirichlet product of two multiplicative functions is multiplicative.

3)show that if $f$ is multiplicative and $f$ is not identically $0$, then $f^{-1}$ is also multiplicative.

4)Deduce that the set of non zero multiplicative functions forms a subgroup of $G$.

This is quite a long question, I know it is getting me to do a step by step guided proof to show that the set of non zero multiplicative functions forms a subgroup of G but wanted to write it all down to avoid confusion.

I know for 1) if $f$ is multiplicative then $f(mn)=f(m)f(n)$ but do not really know where to go from here..

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    How do you define *number theoretic function*? – 2017-02-16
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    If all you are trying to prove is that non-zero multiplicative functions on the integers can't vanish at $1$, just suppose $f(1)=0$ and write $f(n)=f(n\times 1)=f(n)f(1)=0$. I don't really understand what $G$ is, though. – 2017-02-16
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    I guess G is the group of arithmetic functions with Dirichlet convolution.It is actually a ring with addition,I think. – 2017-02-16
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    @BogdanSimeonov That makes sense...and it is true that requiring $f(1)\neq 0$ gets you the group of multiplicative units in that ring. Perhaps the OP can confirm? – 2017-02-16
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    A number theoretic function is a function that assigns a value to each positive integer.. Sometimes known as arithmetic functions. This is a question of several parts. This is the first part, for part 2) I am asked to show that the Dirichlet product of two multiplicative functions is multiplicative. For the third part I am asked to show that if $f$ is multiplicative and $f$ is not identically 0, then $f^{-1}$ is also multiplicative. Finally I am asked to deduce that the set of non zero multiplicative functions forms a subgroup of $G$. – 2017-02-16
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    So I am aware that the first three parts are steps in showing that the set of non zero multiplicative functions forms a subgroup of $G$, I just don't know how to. I am not required to show it is a ring, this goes beyond the scope of what I need. – 2017-02-16

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A number-theoretic (or arithmetic) function is a function $f:\mathbb{N}\rightarrow \mathbb{C}$. Such functions form a factorial ring $D$ under Dirichlet convolution product and pointwise addition $(f+g)(n)=f(n)+g(n)$. The unit group of $D$ consists of all arithmetic functions with $f(1)\neq 0$. The multiplicative arithmetic functions (which satisfy $f(1)=1$) form a subgroup of the group of units of $D$. All this is proved in Tom Apostol's book on analytic number theory in detail.

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    I have edited my question so that you can see what path I'm told to go down. We haven't studied rings at all so it won't be needed for this question. I know that that the set of non zero multiplicative functions will form a subgroup of $G$, but what I'm asking exactly is _how_.to show that using the 4 steps listed above. – 2017-02-16
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    OK. 1.) Because $f(1)=1$ and $1\neq 0$ we have $f\in G$, because $G$ consists of all such functions $f$ with $f(1)\neq 0$. Now continue with 2,3,4. – 2017-02-16
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    Sorry but I can't quite see where $f(1)=1$ comes from? – 2017-02-16
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    Because we have $f(1)=f(1\cdot 1)=f(1)\cdot f(1)$ for multiplicative functions. If $f$ is not zero, then we can cancel by $f(1)$ and obtain $f(1)=1$. – 2017-02-16
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    Thank you, I see it now! so for part 2) would this be right? If the dirichlet product of two multiplicative functions is multiplicative then $(f*g)(mn)$=$(f*g)(m)\cdot(f*g)(n)$. We have $(f*g)(mn)= \sum_{d\mid mn}f(d)g(mn/d)$ = $\sum_{d\mid m}f(d)g(m/d)\cdot \sum_{d\mid n}f(d)g(n/d)$ (don't know if I can just jump straight to this step?) = $(f*g)(m)\cdot(f*g)(n).$ as required – 2017-02-16
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    Almost, just some arguments more. For a complete proof see Theorem 2.14 in Apostol, page $35$. It is a good idea here to consult this book, as I said. It solves 1,2,3,4 very clearly. – 2017-02-16