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Write $B=D+K$, where $D$ has $A$ in each diagonal block and zero elsewhere, while $K$ has $I$ immediately above the diagonal and zero elsewhere. These block matrices satisfy $DK=KD$: either product has $A$ immediately above the diagonal. Also, $K^i$ has $I$ on its $i$th superdiagonal and is zero elsewhere.

The binomial theorem for commuting matrices therefore yieldsOnly $r=i$ can contribute to block $(k,k+i)$, provided that block lies in the matrix. HenceThus ${\beta }_{i,j}=\left(\genfrac{}{}{0ex}{}{j}{i}\right)$, interpreted as an element of the field. The formula remains valid in positive field characteristic: some of these coefficients can then vanish. This possibility matters in the next part.

Since $\lambda I$ commutes with $J_d$, the binomial theorem for commuting matrices and part (i) giveEquivalently, its $k$th superdiagonal is constant with value $\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k}$ for $0\le k<d$, with the convention that this value is zero when $k>n$.

Write $A=I_3+N$, whereThe binomial theorem for commuting matrices applies because $I_{3}N=N{I}_3$. All terms of degree at least three vanish, so for every positive integer $n$,The same expression directly gives $A$ when $n=1$, so no small exponent is excluded.

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 givesThus $\boxed{T(h)=I_3\text{ for all }h\in H}$. 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.

The given unitriangular group of degree three over F3 has $27$ elements, just as $G$ does, so counting alone cannot distinguish them. Instead compare element orders.

Write an element as $I+U$, with $U$ strictly upper triangular of size $3$. Then $U^3=0$. Over the finite field ${\mathbb{F}}_3$, the binomial theorem for commuting matrices givesEvery nonidentity element of $H$ consequently has element order $3$. In contrast, $x\in G$ has element order $9$, by part (b). An isomorphism preserves element order, soEquivalently, the exponent of a finite group is $3$ for $H$ and $9$ for $G$.