For $e\mid n\in\mathbb{N}$, I want to show that $$ \mathbb{Z}/n\mathbb{Z} = \bigcup_{e\mid n} \frac{n}{e}(\mathbb{Z}/e\mathbb{Z})^\times,$$ where $\mathbb{Z}/n\mathbb{Z}$ are the integers modulo $n$, and $(\mathbb{Z}/e\mathbb{Z})^\times$ are the integers modulo $e$ coprime to $e$. We define $(\mathbb{Z}/1\mathbb{Z})^\times=\{0\}$. We know $\mathbb{Z}/n\mathbb{Z}$ is a group under addition, and is a ring with multiplication. Furthermore, $(\mathbb{Z}/e\mathbb{Z})^\times$ is a multiplicative group. We know that $$|\mathbb{Z}/n\mathbb{Z}|=\sum_{e\mid n}| (\mathbb{Z}/e\mathbb{Z})^\times|=\sum_{e\mid n}\phi(e)=n $$
Eg: for $n=12$ we have $$ \mathbb{Z}/12\mathbb{Z}=\{0,\dots,11\} = 12(\mathbb{Z}/1\mathbb{Z})^\times \cup 6(\mathbb{Z}/2\mathbb{Z})^\times \cup 4(\mathbb{Z}/3\mathbb{Z})^\times \cup 3(\mathbb{Z}/4\mathbb{Z})^\times \cup 2(\mathbb{Z}/6\mathbb{Z})^\times \cup 1(\mathbb{Z}/12\mathbb{Z})^\times \\ = 12\{0\} \cup 6\{1\} \cup 4\{1,2\} \cup 3\{1,3\} \cup 2\{1,5\} \cup 1\{1,5,7,11\} \\ = \{0\} \cup \{6\} \cup \{4,8\} \cup \{3,9\} \cup \{2,10\} \cup \{1,5,7,11\} $$
Everything falls nicely into sets with no overlap, but I just can't seem to prove it. I've tried by showing that $\frac{n}{e_1}(\mathbb{Z}/e_1\mathbb{Z})^\times\cap\frac{n}{e_2}(\mathbb{Z}/e_2\mathbb{Z})^\times=\{\}$ but can't seem to wield the maths to prove this intersection is empty.
Any suggestions or ideas are appreciated. Thanks in advance.
You are abusing notation a bit, but ok.
Let $e$ be a divisor of $n$ greater than $1$. If $a\in (\mathbb{Z}/e\mathbb{Z})^{\times}$, with $1\leq a\lt e$, then letting $x=\left(\frac{n}{e}\right)a$ we have $$\gcd(x,n) = \gcd\left(\frac{n}{e}a,n\right)= \frac{n}{e}\gcd(a,e) = \frac{n}{e}.$$
Conversely, if $x$ is such that $1\leq x\lt n$ and $\gcd(x,n) = \frac{n}{e}$, then $\gcd(a,e)=1$ where $x = \left(\frac{n}{e}\right)a$.
Thus, as $e$ ranges over all divisors of $n$ other than $n$ itself, you get each $x$, $1\leq x\lt n$ exactly once, on the basis of its gcd with $n$. If $\gcd(x,n)=d$, then $x$ will occur in the term corresponding to $e=\frac{n}{d}$, as the element corresponding to $a=\frac{x}{d}$.
Finally, when $e=n$ you get the only missing representative, namely $0$.