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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

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  • $\begingroup$ Your changes were not exactly equivalent then, seeing that the whole $F \setminus \left\{0\right\}$ isn't always cyclic (try $F = \mathbb Q$). $\endgroup$ – darij grinberg Feb 10 '15 at 0:05
  • $\begingroup$ The claim in the question's title is false in general: neither $\;\Bbb Q^*\,,\,\,\Bbb R^*\;$ , etc. are cyclic. It is true though if $\;|\Bbb F|<\infty\;$ . $\endgroup$ – Timbuc Feb 10 '15 at 0:05
  • $\begingroup$ Do you mean $F$ is a finite field? $\endgroup$ – Leafar Feb 10 '15 at 0:10
  • $\begingroup$ @Timbuc It's true if and only if $F$ is finite. The multiplicative group of an infinite field is never cyclic: if the field has characteristic $0$ it contains $\mathbb{Q}$; if it has characteristic $p>0$ it contains elements of multiplicative order $p^2-1$ that do not exist in the infinite cyclic group. $\endgroup$ – egreg Feb 10 '15 at 0:11
  • $\begingroup$ Did you mean to write $ab^{-1} \in G$ and not $ab^-1 \in G$, and, later, $c^i$ instead of $c_i$? $\endgroup$ – Dilip Sarwate Feb 10 '15 at 0:12
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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.

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  • $\begingroup$ Why is, in your argumentation, $1+g+g^2+\cdots+g^{p-1}=0$ ? I know it must be so since the elements $1,g,\dots,g^{p-1}$ are all zeros of the polynomial $X^p-1$ and therefore must be the zeros of $X^p-1$. So, $$X^p-1=(X-1)(X-g)\cdots(X-g^{p-1})$$ and expansion of the RHS gives that $\sum g^i$ is the coefficient of $X^{p-1}$ in $X^p-1$ hence must be $0$. However, your argumentation seems to be the other why round. $\endgroup$ – user 59363 Feb 12 '15 at 10:12
  • $\begingroup$ Notice that the sum, $S$, is such that $gS = S$. Now, if $S$ had a multiplicative inverse, this would be impossible, unless $g=1$, but we know that is not the case, so $S$ has no multiplicative inverse, so it must be $0$. As for taking the argument in the other direction, you could, if you have the remainder theorem proven, or something similar. But I wanted to prove it with the fewest results I could, and I think its worth making these sorts of observations. $\endgroup$ – user24142 Feb 13 '15 at 3:26
  • $\begingroup$ @user24124 OK, thanks a lot for clarification. But this way you still have to "fiddle around" with symmetric polynomials. BTW, what is the "remainder theorem". $\endgroup$ – user 59363 Feb 13 '15 at 8:26
  • $\begingroup$ Ok, I think I know what "remainder theorem" means. Once more, thanks for clarification. (Unfortunately I missspelled your name in my previous comment.) $\endgroup$ – user 59363 Feb 13 '15 at 8:48
  • $\begingroup$ The remainder theorem states that if a polynomial $p(x)$ is such that $p(a)=0$, then $p(x) = (x-a)q(x)$. To use this you need to know that $F[x]$ is a Euclidean ring, which may or may not have been covered. Otherwise, you're assuming that the factorisation techniques you learnt in the reals work in a general field, and that's not obivous imo. You don't really need to mess around with symmetric polynomials either, you can write down the coefficients in terms of sums of powers of $g$, and then simplify them. Either way works fine, but if you have Euclidean division, you can just use that. $\endgroup$ – user24142 Feb 13 '15 at 9:57
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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 $p^k$.

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