1

$F$ is a field and and $H$ is finite sub-group of $(F,\cdot)$ ($F$ without the $0_F$).
I need to prove that $H$ is cyclic. I can use this fact - Can we conclude that this group is cyclic?.
(I don't see how can it's helps me to prove what I need to prove).

Id like to get hints and clues...

Thank you so much!!

Community
  • 1
CS1
  • 2,091

1 Answers1

2

Hint: as $F$ is a field, polynomials of degree $m$ in $F[X]$ have at most $m$ different zeros in $F$.

benh
  • 6,695
  • But how the subgroup is close to the polynomials of degree $m$? – CS1 Jan 03 '14 at 08:17
  • Further hint: suppose your group has order $m$, then all its elements must be roots of $X^m - 1$. – Magdiragdag Jan 03 '14 at 08:19
  • @Magdiragdag, right, but what is the connection between this and the sub-group... – CS1 Jan 03 '14 at 08:30
  • it shows a general method: try to reformulate statements about the elements of $F^\times$ into statements about polynomials in $F[X]$. – benh Jan 03 '14 at 08:33
  • @benh, but not every subset of $F$ is subgroup, this what I don't understand, the connection between $F[X]$ and the subgroups. – CS1 Jan 03 '14 at 08:38
  • @YoavFridman We want to use the mentioned fact, i.e. that $|{x : x^m =1 }|\leq m \forall m$. So we are interested in the subset ${x:x^m =1 } \subseteq F$. We don't need the fact, that $F^\times$ forms a group. It is enough to find a corresponding set in terms of $F[X]$. – benh Jan 03 '14 at 08:45
  • @benh, I think I begin to get it, so how we get the subgroup? and how we prove that the subgroup is cyclic? Thank you! – CS1 Jan 03 '14 at 08:57
  • 2
    @YoavFridman If I were you, I would get the subgroup by "Let H be a finite subgroup of $F^\times$". Then I would proceed to show that it is cyclic by saying "we show that $H$ is cyclic by using the fact, that... The mentioned condition in this fact can be proven for $H$ by considering $F[X]$. In $F[X]$, the set of elements ${x \in H : x^m=1}$ is precisely the finite set..." I leave it to you to continue the proof. – benh Jan 03 '14 at 09:02
  • @benh, can you explain me more about this sentence: "we show that H is cyclic by using the fact, that... " (At your comment). – CS1 Jan 03 '14 at 09:39
  • 1
    Just replace "..." with "a group $G$ is cyclic if ${x \in G : x^m=1}\leq m$ is finite for every $m$." I think the proof is now completely given in the comments, just collect the pieces and glue them together :) – benh Jan 03 '14 at 10:55
  • @benh, but how this connected to subgroup of $F^\times$? – CS1 Jan 03 '14 at 16:25
  • @benh, but how can I be sure that this is true for every subgroup? – CS1 Jan 04 '14 at 18:59
  • @benh, a root of a polynomial if $p(c)=1$ at $F^\times$, but where is suppose to be $x^m$? we are talking about $p(c)=1$... Thank you! – CS1 Jan 04 '14 at 21:11
  • 1
    Ok, look: I wanted to point you to the following proof: Let $H \subset F^\times$ be a finite subgroup. We show that $|{x \in H: x^m=1}|=m$ for all $m \in N$ because this implies that $H$ is cyclic by a theorem we know. So let $m \in N$, then the polynomial $f(x) = x^m-1$ has at most $m$ different roots in $F$. Hence, it has at most $m$ different roots in $H$. Thus, ${x : f(x) = 0} = {x : x^m = 1}$ has at most $m$ elements. This is what we wanted to show. – benh Jan 05 '14 at 02:42