= Solution
{c}
Let $A$ be a three-dimensional <vector space> with <basis> $p,q,z$. Define a <bilinear map> $\bullet:A\times A\to A$ by
$$
p\bullet q=z,\qquad q\bullet p=-z,
$$
with all other products of <basis vectors> zero. Equivalently,
$$
(ap+bq+cz)\bullet(a'p+b'q+c'z)=(ab'-ba')z.
$$
Every product belongs to $kz$, and every product with an element of $kz$ is zero. Hence both $(u\bullet v)\bullet w$ and $u\bullet(v\bullet w)$ vanish for every $u,v,w\in A$. This proves <associativity>, so $A$ is an <associative algebra>. The product is an <alternating multilinear map>, and each nested product in the <Jacobi identity> is zero, so the same product is a <Lie bracket>. It is nonzero because $p\bullet q=z\ne0$.
There is no multiplicative <identity element>: any candidate $1_A$ would satisfy $1_A\bullet z=0$, whereas an <identity element> would require $1_A\bullet z=z$. Thus
$$
\boxed{A=kp\oplus kq\oplus kz,\quad p\bullet q=z=-q\bullet p,\quad A^2=kz,\quad A^3=0.}
$$
This is an <associative Lie product>; its underlying <Lie algebra> is the <Heisenberg Lie algebra>.
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