$F$ a field and $G$ finite subset of $F \setminus \{0\}$ with 1 & satisfying $a, b ∈ G$ then $ab^{−1} ∈ G$. Show that $G$ is cyclic

Question

Let $F$ be a field and let $G$ be a finite subset of $F \setminus \{0\}$ containing $1$ and satisfying the condition that if $a, b ∈ G$ then $ab^{−1} ∈ G$. Show that there exists an element $c ∈ G$ such that $G = \{c^i : i ≥ 0\}$.

This is an excercise from the book: The Linear Algebra a Beginning Graduate Student Should Know; Golan

Answer 1

By the classification theorem for finitely generated Abelian groups, $G$ can be written as the direct sum of some number of cyclic groups $G = \bigoplus_i \mathbb{Z}_{q_i}.$ Now, if $\gcd(q_i, q_j) = 1$ then $\mathbb{Z}_{q_i} \oplus \mathbb{Z}_{q_j} \cong \mathbb{Z}_{q_i q_j}.$ So, if the $\{q_i\}$ are all coprime, then $G$ is cyclic and we are done.

So, assume there are $n$ and $m$ such that $n\neq m$ and $\gcd(q_n, q_m)\neq 1.$ Therefore, there is a prime $p$ and elements $g, h \in G$ such that $g^p = h^p = 1$ and there is no $i$ such that $g^i = h.$

Note that for $1\leq k < p$, $$\sum_{i=0}^{p-1} g^i = g\left( \sum_{i=0}^{p-1} g^i\right) = \sum_{i=0}^{p-1} g^{ik} = 0$$ and so, by fiddling around with symmetric polynomials, you can verify that $$f(x) = x^p - 1 = \prod_{i=0}^{p-1}(x - g^i)$$ for all $x\in F.$ But, $f(h) = 0$ and this implies that the RHS of the equation is also $0$, but this implies there are zero divisors in $F$, which is false, and $G$ must be able to be written as a cyclic group.

Answer 2

Let $G$ be a finite subgroup of the multiplicative group $F^\times$ of a (not necessarily finite) field, $\vert G\vert =n$ and let $d\mid n$; then $F^\times$ contains at most one subgroup $H$ of order $d$ since every element $h$ of $H$ must be a solution of $X^d-1=0$ and there are at most $d$ such elements in $F^\times$. Hence, for each divisor $d$ of $\vert G\vert$ the abelian group $G$ can have at most one subgroup of order $d$; this is sufficient for $G$ to be cyclic (for example, by the classicifaction of finite abelian groups).

Edit: one can, in fact, avoid the use of the classification of finite abelian groups. One has to show that, if $G$ has prime power order, say $\vert G\vert= p^k$, then $G$ is cyclic. $G$ consists of the $p^k$ distinct zeros of $X^{p^k}-1$. The elements of $G$ which have order smaller than $p^k$ must have order a divisor of $p^{k-1}$. But these are zeros of $X^{p^{k-1}}-1$ and there are at most $p^{k-1}$ such elements. Hence there must be elements which have order $p^k$.