Linked Questions

21
votes
3answers
29k views

Why is the multiplicative group of a finite field cyclic? [duplicate]

Why is the multiplicative group $(K\smallsetminus\{0\},\cdot)$ of a finite field $(K,+,\cdot)$ always cyclic?
0
votes
2answers
203 views

$F$ 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 [duplicate]

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 $c ∈ G$ ...
3
votes
0answers
174 views

Any finite subgroup of the abelian group $(F-\{0\},\cdot)$ is cyclic? ($F$ a field) [duplicate]

I found this problem: Suppose that $F$ is a field, and that $(F-\{0\},\cdot)$ is an abelian group. Show that if $H$ is a finite subgroup of $F-\{0\}$, then $H$ is cyclic. What I have done is: ...
0
votes
2answers
72 views

showing group $G$ is cyclic if every number $m$ that divides $(G:1)$, there are atmost $m$ elements such that … [duplicate]

The following problem is from J.S. Milne book Let $G$ be a finite Abelian group. If $G$ has at most $m$ elements of order dividing $m$ for each divisor of $m$ of $(G:1)$, show that $G$ is cyclic. ...
2
votes
0answers
104 views

Finite subgroup of the multiplicative group of a field is cyclic [duplicate]

Dummit and Foote's Abstract Algebra contains a proof that a finite subgroup of the multiplicative group of a field, $F$, is cyclic. However the seems unnecessarily complex to me. Certainly their ...
1
vote
0answers
48 views

Easy proof of the existence of a primitive element in a finite field [duplicate]

Is there an "easy" proof of the existence of a primitive element in a finite field? My book uses three lemmas and a theorem to get there, and I can't shake the feeling that there's an easier way of ...
0
votes
0answers
34 views

Reduced multiplicative residue modulo p [duplicate]

I would like if someone could provide or show me where I can find the proof that $\mathbb{Z} _n^*$ is cyclic when n is prime. In particular I'm after a simple proof that involves fermats little ...
49
votes
6answers
17k views

Order of elements is lcm-closed in abelian groups

How can I prove that if $G$ is an Abelian group with elements $a$ and $b$ with orders $m$ and $n$, respectively, then $G$ contains an element whose order is the least common multiple of $m$ and $n$? ...
6
votes
2answers
22k views

Abelian group that is non-cyclic

Construct an abelian group of order 12 that is not cyclic. Can somebody please explain me with examples non-cyclic groups I'm having a hard time understanding.
8
votes
3answers
7k views

Is every finite subgroup of $C^*=$ set of all non-zero complex numbers cyclic?

Is every finite subgroup of $C^*=$ set of all non-zero complex numbers cyclic? I see that the set $A_n=\{z:z^n=1\}$ is a subgroup of $C^{*}$. Any element of $A_n$ is a solution of $z^n=1$.Now the ...
6
votes
1answer
3k views

About the Center of the Special Linear Group $SL(n,F)$

I want to classify the center of the Special Linear Group. I already determined the center for $SL(n,F)$: $$Z(SL(n,F))=\left\{ \lambda I_n:\lambda^n=1 \right\}.$$ I showed that $Z(SL(n,F))$ is itself ...
2
votes
3answers
788 views

Number of elements which are cubes/higher powers in a finite field.

This question is a slight generalization of This Question. How many elements are there in a finite field of order $q$ which are : Squares. Cubes. Higher powers. I mean : How many elements are ...
4
votes
2answers
1k views

Prove that a finite group of complex numbers are the $n$th roots of unity

Prove that if we have a finite group of complex numbers they are the $n$th roots of unity for some finite integer, $n$. I have seen some clunky ways of doing this but is there a very short simple ...
3
votes
2answers
135 views

Can the additive and the multiplicative inverses to an element in $Z_p$ be the same?

Does it contradict the axioms of a field? I think not. If not so there need to be $a \in Z_7$ and so $3+a=0$ and $3\cdot a=1$ but I can not find this $a \in Z_7.$
4
votes
1answer
902 views

Does there always exist an irreducible polynomial of degree $d$ over $\mathbb{Z}/p\mathbb{Z}$? [duplicate]

Let $p$ be a prime and let $d$ be a positive integer. Does there always exist an irreducible (i.e. unfactorable) polynomial of degree $d$ over $\mathbb{Z}/p\mathbb{Z}$?

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