What is the group theory interpretation of $\sum_{d|n} \phi(d) = n$? Let $\phi$ be the Euler totient function. We know that it satisfies 
 $\sum_{d|n} \phi(d) = n$. Since $\phi(d)$ is the order of $(\mathbb{Z}/d \mathbb{Z})^*$, I was wondering if this equation can be interpreted as some statement in terms of group theory on $(\mathbb{Z}/n \mathbb{Z})$ or $(\mathbb{Z}/n \mathbb{Z})^*$.
I would appreciate if anyone could let me know. Thank you very much!

A possible interpretation is the following one: let $\zeta_1,\ldots,\zeta_n$ the $n$-th roots of unity. They form a cyclic group with order $n$. Each root is a primitive $d$-th root of unity for some $d\mid n$. There are $\varphi(d)$ primitive $d$-th roots of unity, since $\varphi(d)$ is the degree of the cyclotomic polynomial $\Phi_d(x)$. It follows that the sum $$\sum_{d\mid n}\varphi(d)$$
accounts for every $n$-th root of unity, hence it equals $n$.
An alternative approach: since $\varphi$ is a multiplicative function (that means that $\gcd(a,b)=1$ implies $\varphi(ab)=\varphi(a)\varphi(b)$) also
$$ f(n) = \sum_{d\mid n}\varphi(d) $$
is a multiplicative function. It is straightforward to check that for every prime $p$ and every exponent $k$ the identity $f(p^k)=p^k$ holds, hence $f(n)=n$ holds for every $n\in\mathbb{N}^*$. 
