multiplicative Euler's $\phi$ function Here is the pdf and array
I am not understanding the proof that $\phi$ is multiplicative function i.e for relatively prime $m,n$ we have $\phi(mn)=\phi(m)\phi(n)$
There were Three Lemmas before the proving original Theorem which I understood
$L_1: a \text{ is prime to } mn \Leftrightarrow(a,m)=1,(a,n)=1$
$L_2: a=qn+r, (r,n)=1 \text{ Then} (a,n)=1$
$L_3:$ If $c$ be the integer and $(a,n)=1$ Then the number of integers in the set $\{c,c+a,c+2a,\dots,c+(n-1)a\}$ that are prime to $n$ is $\phi(n)$
Then they arranged $mn$ integers in $n$ rows and $m$ collumns , I have not understood this arrangement  and based on this arrangements the proof. could any one help me understand?

The best thing you can do, is  trying it yourself on an example..
You arrange the numbers to the rows, since in each row there are $\Phi(m)$ nubers $a$ which you take in account (i.e. $gcd(m,a) = 1$) and the position of such nubers are always the same in tha rows (so they make a column) and the difference of two nubers in the same column is necessarilly a multiple of $m$. Then you do the reasoning of the forth article, which says in fact the same but for columns.
The general idea is: "write the numbers $\leq mn$ in that matrix and clear the columns in which the numbers have a $gcd$ with $m$ greater than $1$ and with the rest do the same for rows; you must have left $\Phi(m)$ rows and $\Phi(n)$ columns, so at the end you finish with $\Phi(m)\Phi(n)$ numbers; the fact that $gcd(m,n)=1$ ensures that when deleting columns you have not deleted some numbers you would want to delete in the second step".
