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Let $n$ be a positive integer, and let $\mathbb F$ be a field of positive characteristic $p$ with $\gcd(n,p) = 1$. Where can I find some proofs that the group of all $n$-th roots of unity (in an algebraic closure of $\mathbb F$) form a cyclic group? Would you be so kind to provide an account of all known ones of which you are aware (together with references thereof)? Thank you in advance.

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When $\gcd(n,p) \neq 1$ they still form a cyclic group! Just not of order $n$ anymore, i.e. when $n=p$ the $n^{th}$ roots of unity is the trivial group. – dinoboy Dec 24 '12 at 9:03
Your request is somewhat odd; what exactly are you trying to do? – Hurkyl Dec 24 '12 at 9:22

There is a very nice, general theorem that solves part of your question:

Theorem: Any finite subgroup of the multiplicative group of any field is cyclic.

You have a short proof of the above here or here, though I'd strongly advice to read the first chapter of "A Course in Arithmetic", by J. P. fact, reading the whole book would be better.

About your question "Would you be so kind to provide an account of all known ones of which you are aware": I have no idea what you mean by "known ones"...known ones what?

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It seems to be a request for known proofs. However, the shortness of the usual proof of that theorem makes it unlikely that there are any interesting other proofs. – Hagen von Eitzen Dec 24 '12 at 10:48
Why doesn't it solve the whole question? – Lior B-S Dec 24 '12 at 12:02
Oh, because his second question, the one I wrote I've no idea what he means and Hagen explained, wasn't addressed...The question about the roots is done in whole, though. – DonAntonio Dec 24 '12 at 14:40

If the finite Abelian group of $n$-th roots of unity were not cyclic, there would be some prime $q$ for which there were at least $q^2$ $q$-th roots of unity. This contradicts the fact they are all roots of $x^q - 1$ which has at most $q$ roots.

Incidentally, the splitting field of $x^n - 1$ is a cyclotomic field in characteristic 0 and a finite field in positive characteristic.

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A primitive root of unity of order m in a field $K$ is an element $z$ of $K$ such that $z^m=1$ and $z^{r !}= 1$ for any positive integer $r < m$. The element $z$ generates the cyclic group $\mu(m)$ of roots of unity of order $m$.

If in $K$ there exists a primitive root of unity of order $m$, then $m$ is relatively prime to the characteristic of $K$. An algebraically closed field contains a primitive root of any order that is relatively prime with its characteristic. If $z$ is a primitive root of order $n$, then for any $k$ that is relatively prime to $n$, the element $z^k$ is also a primitive root. The number of all primitive roots of order $m$ is equal to the value of the Euler function $\phi(m)$ if $\gcd (m, char(K)) = 1$.

Now, what I was saying is that we can write a proof as follows: if there exists a homomorphism from the group formed by the roots to the group $(F,*)$ such that the kernel of the homomorphism contains the group of roots, then the group of roots is necessarily mapped into the multiplicative identity. Since the multiplicative identity generates the group $(F,+)$ (so $(F,+)$ is cyclic) of order $p$ and since the group of roots was obviously cyclic of order $n$, then the group of roots forms a cyclic group in an algebraic closure of F which is of order $p^n$. Like I said, it follows straight from the sylow theorems and the isomorphism theorems.

For the first part I used:

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The only use really for the nth-roots of unity part is to visualize the roots in terms of modular arithmetic. Since gcf(n,p)=1, there exists a cyclic group generated by n of order p(since there exists one generated by p), which is a subgroup of (F,+) since F must be closed. The rest is Trivial, since the roots obviously form a cyclic subgroup by modular arithmetic. The entire proof is a special case of the isomorphism theorems. You could just consider a mapping from the roots to (F,*) such that the nth roots are the kernel of a homomorphism and then show that such a mapping exists. Alternatively, but really along the same line of reasoning you could consider what happens when you add the identity element to itself n times...

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The above really doesn't answer anything the OP asked. You may want to check this carefully, @Cardi. There also several claims which are, in the bext of cases, pretty confusing: what does "Since gcf(n,p)=1, there exists a cyclic group generated by n of order p(since there exists one generated by p)" really mean?? Did you mean the trivial group generated by $p$, since $\,p=0\pmod p\,$ or what? And if you meant this, how is this relevant to anything at all? – DonAntonio Dec 24 '12 at 10:10

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