I study a proof of Hurwitz's theorem on bilinear compositions of sums of squares. The proof relies on the following lemma (which is given as an exercise):
Lemma. Let $G=\langle a_1,\dots,a_{n-1}\mid a_i^2=\epsilon,\epsilon^2=1,a_ia_j=\epsilon a_ja_i \textrm{ for }i\ne j\rangle.$ Then
- $|G|=2^n;$
- $[G,G]=\langle1,\epsilon\rangle;$
- $Z(G)=\{1,\epsilon\}$ for odd $n$ and $Z(G)=\{1,\epsilon,a_1\dots a_{n-1},\epsilon a_1\dots a_{n-1}\}$ for even $n.$
It is clear that $\epsilon\in Z(G),$ so $H=\{1,\epsilon\}$ is normal in $G.$ Since $G/H$ is Abelian, either $[G,G]=\{1\}$ (and, hence, $G$ is Abelian and $\epsilon=1$) or $[G,G]=\{1,\epsilon\}.$
$G/H$ is a direct product of $n-1$ copies of $\mathbb Z_2,$ consequently, $|G/H|=2^{n-1}.$ And to prove the first two statements it's enough to check that for $n>2$ there exists a non-Abelian group with those relations.
Here I'm stuck. Could you please help me?
UPD: I see there's no need to construct such a group to gain a proof but this question (i.e., the existence of this group) has intrigued me on it's own.