= Solution
{c}
Yes: use the <non-abelian group> $H$ from part (a). Every element is $I_3+N$ for a <strictly upper triangular matrix> $N$ with $N^3=0$. In the <finite field> $\mathbb Z_3$, the <binomial theorem for commuting matrices> gives
$$
(I_3+N)^3=I_3+3N+3N^2+N^3=I_3.
$$
Thus $\boxed{T(h)=I_3\text{ for all }h\in H}$. This constant identity-valued map is a <group homomorphism>, since $T(xy)=I_3=T(x)T(y)$. Part (a)(iv) already proves that $H$ is not an <abelian group>. Here $T$ is neither injective nor surjective, consistently with the preceding part.
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