The ring $\mathbb{Z}_{n m}$ is isomorphic to the ring $\mathbb{Z}_n \times \mathbb{Z}_m$ if $\operatorname{gcd}(m, n)=1$ holds. Let us prove the property that the Totient Function $\varphi$ is multiplicative.
Given Proof:
The function $\varphi: \mathbb{Z}_{n m} \longrightarrow \mathbb{Z}_n \times \mathbb{Z}_m$ with $\varphi(\bar{a}):=(\bar{a}, \bar{a})$ is an isomorphism. Therefore, $\varphi$ bijectively maps the unit group $\mathrm{E}\left(\mathbb{Z}_{n m}\right)$ to the unit group $\mathrm{E}\left(\mathbb{Z}_n \times \mathbb{Z}_m\right)$. Now we know: $\mathrm{E}\left(\mathbb{Z}_{n m}\right)=\mathrm{E}\left(\mathbb{Z}_n\right) \times \mathrm{E}\left(\mathbb{Z}_m\right), \varphi(n)=\left|\mathrm{E}\left(\mathbb{Z}_n\right)\right|.$ It follows: $$\varphi(n m)=\left|\mathrm{E}\left(\mathbb{Z}_{n m}\right)\right|=\left|\varphi\left[\mathrm{E}\left(\mathbb{Z}_{n m}\right)\right]\right|=\left|\mathrm{E}\left(\mathbb{Z}_n \times \mathbb{Z}_m\right)\right|=\left|\mathrm{E}\left(\mathbb{Z}_n\right) \times \mathrm{E}\left(\mathbb{Z}_m\right)\right|=\varphi(n) \cdot \varphi(m),$$ proving that $\varphi$ is multiplicative.
My question: I do not really get every part of this proof.
(i) If we say that $\mathrm{E}\left(\mathbb{Z}_{n m}\right)=\mathrm{E}\left(\mathbb{Z}_n\right) \times \mathrm{E}\left(\mathbb{Z}_m\right)$, why can't we just say $$\varphi(n m)=\left|\mathrm{E}\left(\mathbb{Z}_{n m}\right)\right|=\left|\mathrm{E}\left(\mathbb{Z}_n\right) \times \mathrm{E}\left(\mathbb{Z}_m\right)\right|=\varphi(n) \cdot \varphi(m)$$ (ii) What do the steps in between mean, why can we take the totient function of a unit group? Is this the isomorphism that maps the whole group? Why can we then use the multiplicativity of $E$?
There's nothing so fancy as calculating the totient of a group here; there's just a case of horrendous notation. Shockingly, given the use of $\varphi$ as the symbol for the totient function, the author saw fit to also use $\varphi$ to denote the canonical isomorphism $\Bbb Z_{nm}\cong\Bbb Z_n\times\Bbb Z_m$.
So in the line: $$\varphi(nm)=|E(\Bbb Z_{nm})|=|\varphi[E(\Bbb Z_{nm})]|$$The $\varphi$ on the left is the totient, but the $\varphi$ on the right is the group isomorphism.
Other than that, the proof is fine. The other equalities inbetween are to show the full thought process; given an isomorphism of rings $f:R\cong R'$, we know there is a bijection of sets $E(R)\cong f(E(R))$ (given by $f$) and then $f(E(R))=E(R')$ is easy to prove. So we have the equalities $|E(\Bbb Z_{nm})|=|\varphi[E(\Bbb Z_{nm})]|$ and then $|\varphi[E(Z_{nm})|]=|E(Z_n\times\Bbb Z_m)|$. Then you have to check $E(\Bbb Z_n\times\Bbb Z_m)\cong E(\Bbb Z_n)\times E(\Bbb Z_m)$, so there are two more equalities: $|E(\Bbb Z_n\times\Bbb Z_m)|=|E(\Bbb Z_n)\times E(\Bbb Z_m)|=|E(\Bbb Z_n)|\cdot|E(\Bbb Z_m)|$.
I would agree that this is long winded, but there is nothing wrong with showing every step of the calculation, especially not at the more elementary level.