Question

Consider the finite 2-groups containing cyclic subgroup of index 2:

$C_{2^n}$, $C_{2^{n-1}}\times C_2$, $D_{2^n}$, $SD_{2^n}$, $QD_{2^n}$, $Q_{2^n}$.

Can every finite (non-abelian) 2-group be embedded in $H\times K$ where $H$, and $K$ are one of the groups in the list?

Answer 1

Any subgroup of $H \times K$ will have a subgroup of index at most 4 that is a product of two cyclic groups. This comes from intersecting with a fixed choice of cyclic subgroups of index 2 in each factor.

The direct sum of 3 copies of $\mathbb{Z}/8\mathbb{Z}$ is a counterexample, since it doesn't have a subgroup of index at most 4 that is a product of 2 cyclic groups. If you want a nonabelian group, you can take the product of this with quaternions.