Any group of odd square free order is cyclic So this problem is actually mistaken, the condition should really be that a group of order $n$ with $\gcd(n,\varphi(n))=1$, not just being odd square free. Since there exists a group of order 21 which is not abelian, thanks to @NickyHekster. Sorry for the fallacy conclusion. So I posted a new one with the complete problem here:
Any group of order $n$ satisfying $\gcd (n, \varphi(n)) =1$ is cyclic
 A: It is not true, there is a non-cyclic group of order 21.
A: To find the section, use the Sylow theorems. 
You have found $\pi : G \to H \simeq \mathbb Z / p \mathbb Z$, an epimorphism. Now in $G$, since $|G| = p_1\cdots p_s$, without loss of generality assume $|H| = p_s$. By Lagrange's theorem there exists $x \in G$ with $K \overset{def}=\langle x \rangle \simeq \mathbb Z / p \mathbb Z$.
Using the Sylow theorems, $K$ is a Sylow $p$-subgroup since $|G|$ is squarefree. We wish to show that $K \le G$ is the unique subgroup of $G$ of order $p$. Call the number of such subgroups $n_p$. Since $n_p$ divides $|G|/|K| = p_s$ and $n_p = 1 \pmod{p_s}$, $n_p = 1$. That is, $K$ is unique. Any isomorphism $H \to K$ gives rise to a section $H \hookrightarrow G$. Using the expansion of my suggestion by Giorgio Mossa, you're done.
Added : For those who worry that this argument is wrong, it is not. This is because the problem does not lie here. Using this section, one gets a normal subgroup $N \trianglelefteq G$ and a subgroup $H \le G$ such that $NH = G$ and $N \cap H = \{e\}$, i.e. one gets a semi-direct product. Therefore the most general result one can get is a decomposition
$$
G \simeq (\cdots(( H_1 \rtimes H_2 ) \rtimes H_3 ) \rtimes \dots  \rtimes H_s
$$
where $|H_i| = p_i$, using induction on $i$ up to $s$ where $|G| = p_1 \cdots p_s$.
Hope that helps,
A: Here are some hint that can help you in proving the part b).
When you have a surjective homomorphism 
$$\pi \colon G \to H$$
and want to show that $G \cong H \times N$ for $N=\ker \pi$ a standard technique 
is to find a morphisms $i \colon H \to G$ such that $\pi \circ i=1_H$, that's the section which Patrick Da Silva was referring to.
If you find such a map then clearly


*

*$i$ must be injective, since $\pi\circ i=1_H$ and $\ker i \subseteq \ker \pi\circ i= 0$;

*there's a subgroup $H'=\text{Im } i < G$ isomorphic to $H$, because $i$ is injective;

*the intersection $H' \cap N=(0)$, because $\pi \circ i=1_H$ there cannot be any element in $\text{Im }i$ which is element of $N$


from that you should easily get that $G$ is a semidirect product of $H$ and $N$, to have a direct product you just need to prove that $H'$ is a normal subgroup of $G$ (hint: $H'$ have order $p$ for some prime that divide the order of $G$, and is the maximal power of $p$ dividing it). 
The last part is to prove that $N$ is an abelian group, in order to prove that $G$ is abelian (hint: induction).
Hope this helps.
