Are All Multiplicative Group Cyclic?

For a prime number p, the group (Z/pZ)^× is always cyclic, consisting of the non-zero elements of the finite field of order p. More generally, every finite subgroup of the multiplicative group of any field is cyclic.

Is every subgroup of a cyclic group cyclic?

Theorem: All subgroups of a cyclic group are cyclic. If G=⟨a⟩ is cyclic, then for every divisor d of |G| there exists exactly one subgroup of order d which may be generated by a|G|/d a | G | / d . Proof: Let |G|=dn | G | = d n .

Why is the multiplicative group of a finite field cyclic?

Theorem 3.11. Every finite subgroup of the multiplicative group of a field is cyclic. Proof. ... Every finite abelian group contains an element of order equal to its exponent, so G contains an element of order m = n = #G and is therefore cyclic.