Group theoretical proof of $\varphi(rs)=\varphi(r)\varphi(s)$ through generators of the group.

Question

Given a group $G=\langle a\rangle$ of order $rs$, with $(r,s)=1$, I showed there exist unique $b,c\in G$ such that $a=bc$ with $b$ of order $r$ and $c$ of order $s$. The latter is a direct consecuense of Bezout's Identity: $a=a^1=a^{xr+ys}$, for $x,y$ unique, so $b=a^r,c=a^s$.

Using the above I am to prove that $\varphi(rs)=\varphi(r)\varphi(s)$; i.e. that Euler's function is multiplicative.

I know that a group of order $n$ has exactly $\varphi(n)$ generators, I just can't relate $\varphi(rs)$ to the generators of $\langle b \rangle$ and $\langle c \rangle$.

Any help would be appreciated.

Answer 1

Since $(r,s) = 1$, $\Bbb Z_{rs} \cong \Bbb Z_r \times \Bbb Z_s$. Show that this implies $\Bbb Z_{rs}^\times \cong \Bbb Z_r^\times \times \Bbb Z_s^\times$ and therefore $\varphi(rs) = \varphi(r) \varphi(s)$. ($\Bbb Z_n^\times$ is the group of units of the cyclic group of order $n$.)

Answer 2

Show that your group $(a)$ is isomorphic to the direct product of $(b)$ and $(c)$.

Next, count the number of generators in the group $(b)\times(c)$.