If $F$ is any field, there is a subgroup of $\text{PGL}_2(F)$ given by fractional linear transformation of the form $z \mapsto az + b$ where $a \in F^{\times}, b \in F$. This is precisely the subgroup fixing the point at infinity in $\mathbb{P}^1(F)$. Taking $F$ to be a finite field $\mathbb{F}_q$ we obtain a family of (usually) nonabelian Frobenius groups of order $q(q - 1)$.
Now, we can further restrict $a$ to lie in any subgroup of $F^{\times}$; in particular, for $q$ odd, taking it to lie in the subgroup of quadratic residues gives a family of (usually) nonabelian groups of order $\frac{q(q-1)}{2}$. Taking $q = 7$ gives the desired group.
Yes this holds for Fermat primes. You need the following observation. This is sometimes called the $N/C$ Theorem.
Lemma Let $G$ be a group with a subgroup $H$, then $N_G(H)/C_G(H)$ embeds homomorphically into Aut$(H)$.
Here $N_G(H)=\{g \in G : gH=Hg\}$, the normalizer of $H$ in $G$ and $C_G(H)=\{g \in G : gh=hg \text { for all } h\in H\}$, the centralizer of $H$ in $G$.
Now let us apply this to the situation where $N=H$ is normal, $|N|=$ Fermat-prime and $|G|$ is odd. Then $N_G(N)=G$ because of the normality of $N$. And since $N$ is cyclic, |Aut$(N)|$ is a power of $2$ (in fact $|N|-1$). It follows from the $N/C$ theorem that $|G/C_G(N)|$ is a power of $2$. But obviously it also divides $|G|$, which is odd. This can only be when $G=C_G(N)$, that is $N \subseteq Z(G)$.
So how do you prove the lemma? Let me give a sketch and leave the details with you: $N_G(H)$ acts as automorphisms on $H$ by conjugation. The kernel of the action is $C_G(H)$.
Generalizations:
Proposition 1 Let $G$ be a finite group, $N$ a normal subgroup of prime order $p$, with gcd$(|G|,p-1)=1$, then $N \subseteq Z(G)$.
Note that this holds for $p=2$: a normal subgroup of order $2$ must be central.
Proposition 2 Let $G$ be a finite group, $N$ a normal subgroup of prime order $p$, where $p$ is the smallest prime dividing $|G|$. Then $N \subseteq Z(G)$.
Best Answer
Let $\,G\,$ be a group, $\,G'=[G,G]=\,$ its derived or commutator subgroup, then we have the following:
(1) $\,G/G'\,$ is abelian, and thus
(2) Any product in $\,G/G'\,$ can be arranged at will by (1), and finally
(3) The product of all the elements in a group with an odd number of elements is in $\,G'\,$ since the product of their images in $\,G/G'\,$ is trivial.