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QUESTION

Prove the following facts about the Dirichlet product $f*g$ of
arithmetic functions $f$ and $g$:-

\begin{description}

\item[(i)]
If $f$ and $g$ are multiplicative, so is $f*g$.

\item[(ii)]
$f*g=g*f$.

\item[(iii)]
$(f*g)*h=f*(g*h)$.

\end{description}



ANSWER

\begin{description}

\item[(i)]
Suppose gcd$(m,n)=1$. We need to prove
$(f*g)(mn)=(f*g)(m)(f*g)(n)$. to see how to perform the
manipulations, it helps to write both down:-

$$(f*g)(mn)=\sum_{d|mn}f(d)g\left(\frac{mn}{d}\right)$$ and
$$(f*g)(m)(f*g)(n)=\sum_{d|m}f(d)g(\left(\frac{m}{d}\right)
\sum_{d|n}f(d)g\left(\frac{n}{d}\right)$$.

It is probably easier to manipulate the second expression to get
the first. First we will use different symbols for the divisors of
$m$ and the divisors of $n$ (at present $d$ is used for both ) and
then we'll combine the sums:-

\begin{eqnarray*}
(f*g)(m)(f*g)(n)&=&\sum_{d_1|m}f(d_1)g\left(\frac{m}{d_1}\right)
\sum_{d_2|n}f(d_2)g\left(\frac{n}{d_2}\right)\\
&=&\sum_{d_1|m,d_2|n}f(d_1)g\left(\frac{m}{d_1}\right)f(d_2)g
\left(\frac{n}{d_2}\right)\\&=&\sum_{d_1|m,d_2|n}f(d_1)f(d_2)g\left(\frac{m}{d_1}\right)
g\left(\frac{n}{d_2}\right).
\end{eqnarray*}

Now gcd$(m,n)=1$, so if $d_1|m$ and $d_2|n$ then gcd$(d_1,d_2)=1$
and gcd$\left(\frac{m}{d_1},\frac{n}{d_2}\right)=1$, so our sum
becomes
$\sum_{d_1|m,d_2|n}f(d_1d_2)g\left(\frac{mn}{d_1d_2}\right)$.

Now noting that as $d_1$ ranges over the divisors $m$ and $d_2$
over the divisors of $n, d_1d_2$ ranges over the divisors of $mn$
(since gcd$(m,n)=1)$, we see that this is
$\sum_{d|mn}f(d)g\left(\frac{mn}{d}\right)=(f*g)(mn)$ as required.


\item[(ii)]

\begin{eqnarray*}
f*g(n)&=&\sum_{d|n}f(d)g\left(\frac{n}{d}\right)\\
&=&\sum_{d_1d_2=n}f(d_1)g(d_2)\\ &=&\sum_{d_2d_1=n}f(d_1)g(d_2)\\
&=&\sum_{d_1d_2=n}f(d_2)g(d_1)\\
&=&\sum_{d|n}f\left(\frac{n}{d}\right)g(d)=g*f(n).
\end{eqnarray*}

\item[(iii)]

\begin{eqnarray*}
((f*g)*h)(n)&=&\sum_{d|n}(f*g)(d)h\left(\frac{n}{d}\right)\\
&=&\sum_{d_1d_2=n}(f*g)(d_1)h(d_2)\\
&=&\sum_{d_1d_2=n}\left(\sum_{d|d_1}f(d)g\left(\frac{d_1}{d}\right)\right)h(d_2)\\
&=&\sum_{d_1d_2=n}\left(\sum_{de=d_1}f(d)g(e)\right)h(d_2)\\
&=&\sum_{ded_d=n}f(d)g(e)h(d_2)\\
&=&\sum_{d_1d_2d_3=n}f(d_1)g(d_2)h(d_3).
\end{eqnarray*}

Similarly

\begin{eqnarray*}
(f*(g*h))(n)&=&\sum_{d|n}f(d)(g*h)\left(\frac{n}{d}\right)\\
&=&\sum_{d_1d_2=n}f(d_1)(g*h)(d_2)\\
&=&\sum_{d_1d_2=n}f(d_1)\left(\sum_{d|d_2}g(d)h\left(\frac{d_2}{d}\right)\right)\\
&=&\sum_{d_1d_2=n}f(d_1)\left(\sum_{de=d_2}g(d)h(e)\right)\\
&=&\sum_{d_1de=n}f(d_1)g(d)h(e)\\
&=&\sum_{d_1d_2d_3=n}f(d_1)g(d_2)h(d_3).
\end{eqnarray*}

These are the same, so $((f*g)*h)=(f*(g*h))$ as required.

\end{description}




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