Finite 2-groups with $(ab)^{2}=(ba)^{2}$

Question

There exist nonabelian finite 2-groups $G$ with the property $(A2)$ : for every $a,b\in G$, $(ab)^{2}=(ba)^{2}$. An example of a such group is given by the quaternion group $Q_{8}$ of order 8. Is there any classification of such groups ?

Answer 1

The first observation is that the given condition is equivalent to all squares being central, because substituting $c=ab$ in the relation turns it into $c^2=bc^2b^{-1}$. Next, to explain Derek Holt's comment, for all $a$ and $b$ we have $$(abab)^2=abab.baba=abaaba.b^2=\dotsb=a^4b^4,$$ so $$(aba^{-1}b^{-1})^2=(abaa^{-2}bb^{-2})^2=(abab)^2a^{-4}b^{-4}=1.$$ It follows that all commutators square to the identity. What's more, in any group a commutator is a product of three squares, so in your group commutators are central.

So there's a universal group $H$ on $n$ generators, and of exponent $2^m$, of this form. If the generators are $g_1,\dots,g_n$ then the relations say that each $g_i^2$ is central and has order $2^{m-1}$, and each $[g_i,g_j]$ is central of order two. This group $H$ sits in an exact sequence $$1\to (\mathbb{Z}/2^{m-1})^n \times (\mathbb{Z}/2)^{\binom{n}{2}}\to H \to (\mathbb{Z}/2)^n \to 1$$ where the normal subgroup on the left is $\Phi(H)$ and is central. Any $2$-group $G$ satisfying your condition, with $|G/\Phi(G)|=2^n$ and exponent dividing $2^m$ can be obtained as a quotient of this $H$ by a subgroup of $\Phi(H)$.