I would like to know if this reasoning is sound. This comes from exercise 2.21 in Rotman's intro to group theory. The goal is to show $\phi(rs)=\phi(r)\phi(s)$.
I was able to show that if $G$ is cyclic of order $n$, then $a^k$ , an element of $G$, is a generator iff $k$ and $n$ are relatively prime: thus there are $\phi(n)$ generators of $G$. (I should mention $\phi(n)$ = the number of integers $x$ where $1\leq x< n$ such that $x$ and $n$ are relatively prime). Next, that if the order of $G$ is $rs$, where $r$ and $s$ are relatively prime, then there are unique subgroups of $G$ of order $r$ and $s$. These subgroups are those generated by $b$ and $c$, where $b=a^r$ , $c=a^s$ , the order of $b$ is $s$, and the order of $c$ is $r$. Further, $bc=a^{(r+s)}$ is a generator of $G$ (since if $r$ and $s$ are relatively prime, then $r+s$ is relatively prime with $rs$). Everything up until now I'm fine with, the following is what I want to prove. Now $b$ has $\phi(s)$ many generators, and $c$ has $\phi(r)$ many. That means there are $\phi(s)$ many integers $x$ such that $a$ is in $G$ and $a^x = b$, and $\phi(r)$ many integers $y$ such that $a^y = c$. So ostensibly there are $\phi(r)\phi(s)$ combinations $a^{(x+y)} = bc$, and thus there are $\phi(r)\phi(s)$ generators of $G$. But we know there are $\phi(rs)$ generators of $G$, so $\phi(rs)=\phi(r)\phi(s)$