For which $n$, is $\Bbb{Z}_n \cong H \oplus (\Bbb{Z}_n/H) $ is true? 
Give a characterization of all positive integers $n$ for which $$\Bbb{Z}_n \cong H \oplus (\Bbb{Z}_n/H) $$ for every subgroup $H$ of $\Bbb{Z}_n$. 

For $n$ is prime ,then $\Bbb{Z}_n $ has only two improper subgroups. So the conclusion is true. What about other $n$'s?
 A: There are a few facts that need to be known to finish this exercise: 
(1) a subgroup of a cyclic group is cyclic. 
This is an easy exercise, either by lifting the subgroup to $\mathbb{Z}$ or by considering a given generator $g$ of the big group and the minimal power $n$ such that $g^n \in H$
(2) A quotient of a cyclic group is cyclic. 
Again, this is an easy exercise: the image of a generator under the quotient map is a generator of the quotient group. 
(3) The chinese remainder theorem: $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$ is cyclic (hence of order $mn$) if and only if $m\land n = 1$. 
This is a classical result.
(4) For any $d\mid n$, there is a (unique- but it's not necessary for the remainder of the proof) subgroup of $\mathbb{Z}/n\mathbb{Z}$ of order $d$. 
Put $H:=\langle \frac{n}{d}\rangle$. 
With these in mind, if $n$ is such that for all $H\leq \mathbb{Z}/n\mathbb{Z}$, $\mathbb{Z}/n\mathbb{Z}\simeq H\oplus ((\mathbb{Z}/n\mathbb{Z})/H)$; then this means, by (1),(2) and (3), that for all subgroups $H$ of $\mathbb{Z}/n\mathbb{Z}$, $|H|$ and $\frac{n}{|H|}$ are coprime, and by (1),(2) and (3), this condition is also sufficient. 
By (4) and Lagrange's theorem, this is equivalent to saying that for all $d\mid n$, $d\land \frac{n}{d} =1$. 
If $n=p_1...p_k$ with $p_i$'s distinct primes, then clearly this is the case. If $p^2\mid n$, $p$ prime then $p\mid p \land \frac{n}{p}$, and so this isn't the case. 
In conclusion, this is true of $n$ if and only if $n$ is a product of distinct primes (aka $n$ is squarefree)
