Euler's formula and subgroups of $\mathbb Z_n$

Question

Prove that in $\mathbb Z/n\mathbb Z$ for every divisor $d$ of $n$ there is a unique subgroup of order $d$ using the following results: $\sum_{d\mid n}\varphi(d)=n$ and the number of generators of $\mathbb Z/n\mathbb Z$ is $\varphi(n)$.

Answer 1

Let $A_d$ be the set of elements of order $d$ in $\mathbb Z/n\mathbb Z$, $d\mid n$. Obviously $A_d\neq\emptyset$ (residue class of $n/d$ has order $d$) and then $|A_d|\ge\varphi(d)$ (the number of elements of order $d$, or the number of generators, in the subgroup generated by the residue class of $n/d$). Since $(A_d)_{d\mid n}$ is a partition of $\mathbb Z/n\mathbb Z$, it follows that $\sum_{d\mid n}|A_d|=n$ and therefore (from Euler's formula) $|A_d|=\varphi(d)$ for all $d\mid n$. The residue class of $n/d$ generates a subgroup of order $d$ and this subgroup has exactly $\varphi(d)$ elements of order $d$. It follows that this is the only subgroup of $\mathbb Z/n\mathbb Z$ with $d$ elements.