All 2-groups have cyclic commutator subgroups?

Question

I have seen that groups of order 8 and order 16 have commutator subgroup cyclic. I want to know if that can be extended to all 2-groups or if there is a reason why this happens.

Answer 1

No: if $G$ and $H$ are groups, then $[G\times H,G\times H]= [G,G]\times[H,H]$. So all you need to do is take the direct products of two $2$-groups with nontrivial commutator subgroup to immediately get one with noncyclic commutator subgroup. In particular, $D_8\times D_8$ (where $D_8$ is the dihedral group of order $8$) is an example.

What you noticed is a consequence of one might call the "law of small groups": there aren't enough groups of small order to avoid coincidences. Any nonabelian group of order $p^3$ has cyclic commutator subgroup (equal to the center) of order $p$. Something similar happens with groups of order $p^4$. There just isn't "enough room" for a noncyclic commutator subgroup. Once you get to $p^6$ you can use the easy trick I start with above.

But you also get examples with order $p^5$. Take the group of all $4\times 4$ upper triangular unipotent matrices and entries in $\mathbb{F}_p$, of the form $$ \left(\begin{array}{cccc} 1 & 0 & * & *\\ 0 & 1 & * & *\\ 0 & 0 & 1 & *\\ 0 & 0 & 0 & 1 \end{array}\right).$$ Here the commutator subgroup is isomorphic to $C_p\times C_p$. The group is generated by $3$ elements, two of which commute with each other but not with the third generator.