Example of a group with certain properties. I'm trying to find an example of a group $G$ such that $K\lhd H\lhd G$ are normal subgroups of $G$. Further $H/K$ is not a subgroup of any homomorphic image of $G$.
 A: The example with $|G|$ minimal have $|G|=64$. The examples with $|G|$ of order 64 have $G$ isomorphic to SmallGroup(64,k) for k in [18,25,33,35,36].
I didn't see any examples of order 64 with a particularly simple description.
The examples with $|G|$ minimal subject to $|G|>64$ have order 96, and $G$ isomorphic to SmallGroup(96,k) for k in [3,13,44,71].
These are the only examples with $G$ of order less than 128.
Here is one example written out with a computer-free proof (though I do assume familiarity with Sylow theorems and $p$-groups).
SmallGroup(96,71)
$$G=\langle a,x,y : a^6 = x^4 = y^4 = 1, xa = ay^{-1}, ya=ax^{-1}y, yx=xy \rangle$$
$$
a=\begin{bmatrix}0&-1&0\\-1&1&0\\0&0&1\end{bmatrix}, \quad
x=\begin{bmatrix}1&0&0\\0&1&0\\1&0&1\end{bmatrix}, \quad
y=\begin{bmatrix}1&0&0\\0&1&0\\0&1&1\end{bmatrix}, \quad \mod 4
$$
$$N=\langle a^3,x,y \rangle, \quad K =\langle x^2 \rangle$$
Note that $N/K$ has order 16, but that the only normal subgroup of $G$ with index a multiple of 16 is $1$ ($G$ has normal subgroups of index 1,2,3,6,12,24,96). We can prove this without too much trouble, as $G$ clearly has a unique minimal normal subgroup, $\langle x^2, y^2 \rangle$, so that any non-identity normal subgroup has order divisible by 4, and thus index dividing $96/4 = 24$.
It thus suffices to show that $N/K$ is not isomorphic to a subgroup of $G$. Since $N$ is a Sylow 2-subgroup of $G$ and $N/K$ is a 2-group, it suffices to show $N/K$ is not isomorphic to a subgroup of $N$. If it were, then it would be a maximal subgroup, as it would have index 2. However $N/K$ is not generated by two elements, while every maximal subgroup of $N$ is generated by two elements. $N/K$ is not generated by two elements since $\Phi(N/K) = \Phi(N)/K$ and $\Phi(N) = \langle x^2, y^2 \rangle$. A maximal subgroup $M$ of $N$ is two-generated, because (a) it is a maximal subgroup $N/\Phi(N)$, (b) $N/\Phi(N)$ is a three dimensional vector space over $\mathbb{Z}/2\mathbb{Z}$, so (c) every basis of a hyper-plane in $(\mathbb{Z}/2\mathbb{Z})^3$ has two elements, (d) $M/\Phi(N)$ is such a hyperplane, (e) $\Phi(M) = \Phi(N)$ since either $a$ and an element of $\langle x,y\rangle$ generate it with commutators, or $M=\langle x,y\rangle$ so that $x^2,y^2 \in M^2 \leq \Phi(M)$, so (f) $M$ is also two-generated.
Alternatively, this is a very finite calculation.
