= Solution
{c}
Write $q=|\mathbb F|$. The <unit group> $U(\mathbb F)=\mathbb F\setminus\{0\}$ is a <finite abelian group> of <group order> $q-1$, since <multiplication> in a <field> satisfies <commutativity>. The <structure theorem for finitely generated abelian groups>, applied to this finite <group>, gives <invariant factors of a finitely generated module>
$$
U(\mathbb F)\cong\mathbb Z_{d_1}\oplus\cdots\oplus\mathbb Z_{d_r},\qquad
1<d_1\mid d_2\mid\cdots\mid d_r,
$$
after omitting trivial factors. If there are no factors, the <unit group> is trivial and hence a <cyclic group>.
Otherwise put $D=d_r$. Every element $u\in U(\mathbb F)$ satisfies $u^D=1$, because the <order of a group element> in each coordinate divides $D$. Hence all $q-1$ distinct nonzero elements of $\mathbb F$ are <roots of a polynomial> $X^D-1$. The <root bound for a polynomial> over a <field> gives
$$
q-1\leq D.
$$
For completeness, that bound follows by induction from the <factor theorem>: if a nonzero <polynomial> $f$ vanishes at $a$, write $f=(X-a)g$; every other <root of a polynomial> is a root of $g$, because a <field> has no <zero divisors>. Each new distinct root therefore uses one unit of <polynomial degree>.
On the other hand, the <group order> in the decomposition is
$$
q-1=d_1\cdots d_r\geq d_r=D.
$$
Equality is forced, so $d_1\cdots d_{r-1}=1$. As every retained factor is greater than $1$, there can be only one. Consequently
$$
\boxed{U(\mathbb F)\cong\mathbb Z_{q-1}\text{ is cyclic}}.
$$
The same proof establishes the more general result <finite multiplicative subgroup of a field is cyclic>. The <square-free integer> result in part (c)(i) alone would not suffice, since $q-1$ need not be a <square-free integer>; the <root bound for a polynomial> rules out multiple <invariant factors of a finitely generated module> even then.
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