Non-Abelian group of order 64 with an order-4 quotient group the Klein four-group Are there some examples of the non-Abelian group of order 64 say $G$ with a Klein four-group  ($\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} $) quotient group? What are these examples? (Incomplete and partial examples are OKAY.) What is the corresponding normal subgroup $N$ then? For $Q/N=\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} $?
 A: Suppose you have a direct product of groups, say $G = A \times B$.  You can check that the standard projection maps $\pi_1:G \rightarrow A$ and $\pi_2:G \rightarrow B$ defined by $\pi_1(a, b) = a$ and $\pi_2(a, b) = b$ are group homomorphisms.  It's also easy to check that $\ker(\pi_1) \cong B$ and $\ker(\pi_2) \cong A$.
Applying this fact to the problem at hand, we just need to find a nonabelian group of order that divides $64/| \mathbb{Z}_2 \times \mathbb{Z}_2| = 16$.  The quaternion group of order $8$ fits the bill.  So by letting $G$ be an appropriate direct product, we can arrive at an example you're looking for.
A: One way of generating examples of non-abelian groups of order a power of two is to consider upper triangular matrices over the field of two elements $F=\Bbb{Z}/2\Bbb{Z}$. If we let $G$ be the set of such 4x4 matrices
$$
G=\left\{\left(\begin{array}{cccc}1&a&b&c\\0&1&d&e\\0&0&1&f\\0&0&0&1\end{array}\right)\bigg\vert\,a,b,c,d,e,f\in K\right\},
$$
then we clearly get a group (a subgroup of $GL_4(F)$. There are six variables ranging over $F$, so there are $64$ elements in $G$. The subsets where certain entries are set to zero form normal subgroups (IIRC if we set some position to zero, we also need to do the same between the entries of the diagonal and that position). So for example if we define $H$ to be the set of elements of $G$ such that $a=d=0$, then $H$ will be a normal subgroup of $G$. Furthermore, it is
easy to see that $G/H$ is the Klein 4-group because the multiplication of cosets of $H$ amounts to addition of pairs $(a,d)$ in $F^2\simeq V_4$.
