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Any finite field of $q$ elements has exactly $Phi(q-1)$ primitive roots

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$begingroup$

Is the following prove of the above statement correct?

$bullet $Any finite field of $q$ elements is isomorphic to $mathbbF_q$ and we know that $mathbbF_q^*$ is a cyclic group of $q-1$ elements. Let $mathbbF_q^*=langle alpha rangle$ with $alphain mathbbF_q^*$. If we take another $betainmathbbF_q^*$, we can therefore write it as a power of $alpha$: $beta=alpha^k$ for some $k$.

Now we have that $mathbbF_q^*=langle betarangle iff gcd(q-1,k)=1$ as we know that $mathbbF_q^*$ is cyclic of order $q-1$, thus isomorphic to $mathbbZ/(q-1)mathbbZ$.

This in turn gives us $gcd(q-1,k)=1 iff kin (mathbbZ/(q-1)mathbbZ)^*$ and the latter group has $Phi(q-1)$ elements, where $Phi$ denotes the Euler totient function.

Therefore, there are $Phi(q-1)$ choices for $k$, so $Phi(q-1)$ choices for $beta$ such that $mathbbF_q^*=langle beta rangle$. In other words: there are $Phi(q-1)$ primitive roots in each field with $q$ elements. $bulletbullet$

I am particularly unsure of the step where we conclude that $gcd(q-1,k)=1$. It seems slightly logical, in a simpler group setting of $mathbbZ/nmathbbZ$, but I can't quite understand or proof exactly why that particular statement is true.

field-theory finite-fields primitive-roots

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asked Mar 22 at 15:21

MarcMarc

510211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A cyclic group of order $m$ has $varphi(m)$ generators by a simple argument. Now use the fact that $mathbbF_q^*$ is cyclic. It's easier to think about this as $mathbbZ_q-1$ additively, which it is up to (unknown) iso. (This is basically what you're doing.)
  $endgroup$
  – Randall
  Mar 22 at 15:53
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Is the following prove of the above statement correct?

$bullet $Any finite field of $q$ elements is isomorphic to $mathbbF_q$ and we know that $mathbbF_q^*$ is a cyclic group of $q-1$ elements. Let $mathbbF_q^*=langle alpha rangle$ with $alphain mathbbF_q^*$. If we take another $betainmathbbF_q^*$, we can therefore write it as a power of $alpha$: $beta=alpha^k$ for some $k$.

Now we have that $mathbbF_q^*=langle betarangle iff gcd(q-1,k)=1$ as we know that $mathbbF_q^*$ is cyclic of order $q-1$, thus isomorphic to $mathbbZ/(q-1)mathbbZ$.

This in turn gives us $gcd(q-1,k)=1 iff kin (mathbbZ/(q-1)mathbbZ)^*$ and the latter group has $Phi(q-1)$ elements, where $Phi$ denotes the Euler totient function.

Therefore, there are $Phi(q-1)$ choices for $k$, so $Phi(q-1)$ choices for $beta$ such that $mathbbF_q^*=langle beta rangle$. In other words: there are $Phi(q-1)$ primitive roots in each field with $q$ elements. $bulletbullet$

I am particularly unsure of the step where we conclude that $gcd(q-1,k)=1$. It seems slightly logical, in a simpler group setting of $mathbbZ/nmathbbZ$, but I can't quite understand or proof exactly why that particular statement is true.

field-theory finite-fields primitive-roots

share|cite|improve this question

asked Mar 22 at 15:21

MarcMarc

510211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A cyclic group of order $m$ has $varphi(m)$ generators by a simple argument. Now use the fact that $mathbbF_q^*$ is cyclic. It's easier to think about this as $mathbbZ_q-1$ additively, which it is up to (unknown) iso. (This is basically what you're doing.)
  $endgroup$
  – Randall
  Mar 22 at 15:53
  
  

  
  
  
  

add a comment | 

$begingroup$

Is the following prove of the above statement correct?

$bullet $Any finite field of $q$ elements is isomorphic to $mathbbF_q$ and we know that $mathbbF_q^*$ is a cyclic group of $q-1$ elements. Let $mathbbF_q^*=langle alpha rangle$ with $alphain mathbbF_q^*$. If we take another $betainmathbbF_q^*$, we can therefore write it as a power of $alpha$: $beta=alpha^k$ for some $k$.

Now we have that $mathbbF_q^*=langle betarangle iff gcd(q-1,k)=1$ as we know that $mathbbF_q^*$ is cyclic of order $q-1$, thus isomorphic to $mathbbZ/(q-1)mathbbZ$.

This in turn gives us $gcd(q-1,k)=1 iff kin (mathbbZ/(q-1)mathbbZ)^*$ and the latter group has $Phi(q-1)$ elements, where $Phi$ denotes the Euler totient function.

Therefore, there are $Phi(q-1)$ choices for $k$, so $Phi(q-1)$ choices for $beta$ such that $mathbbF_q^*=langle beta rangle$. In other words: there are $Phi(q-1)$ primitive roots in each field with $q$ elements. $bulletbullet$

I am particularly unsure of the step where we conclude that $gcd(q-1,k)=1$. It seems slightly logical, in a simpler group setting of $mathbbZ/nmathbbZ$, but I can't quite understand or proof exactly why that particular statement is true.

field-theory finite-fields primitive-roots

share|cite|improve this question

asked Mar 22 at 15:21

MarcMarc

510211

$endgroup$

share|cite|improve this question

asked Mar 22 at 15:21

MarcMarc

510211

share|cite|improve this question

asked Mar 22 at 15:21

MarcMarc

510211

asked Mar 22 at 15:21

MarcMarc

510211

asked Mar 22 at 15:21

MarcMarc

510211