 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Take $x=h(1,0,0)=I_3+E_{12}$ and $y=h(0,0,1)=I_3+E_{23}$. Their <matrix products> are
$$
xy=\begin{pmatrix}1&1&1\\0&1&1\\0&0&1\end{pmatrix},
\qquad
yx=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}.
$$
They differ in the $(1,3)$ entry because $E_{12}E_{23}=E_{13}$ whereas $E_{23}E_{12}=0$. Therefore $xy\ne yx$, and $\boxed{H\text{ is a non-abelian group}}$.

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