Let $r$ be primitive root mod $p$. When $x$ goes from $1$ to $p-1$, then $r^x$ (mod $p$) goes through all the...












2














I'm trying to understand this situation. Why do the powers of primitive roots smaller than $p-1$ generate all DISTINCT elements in $mathbb{Z}_p$? I am aware about what Fermat's little theorem states and about the fact that $mathbb{Z}_p$ has $p-1$ elements in it. Just not sure how to prove that those elements are distinct and I'm having a struggle with finding proper texts about this.






abstract-algebra number-theory elementary-number-theory cryptography







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  • We define primitive root of $p$ to be a number whose order is $p-1$. Then you should be asking why/how at least one primitive root exists for every prime number..
    – rsadhvika
    Dec 8 at 8:17












  • You can try David Burton's Elementary Number Theory.
    – Brahadeesh
    Dec 8 at 8:25










  • Thank you, I will definetly check this one.
    – R. Jacks
    Dec 8 at 8:37
















2














I'm trying to understand this situation. Why do the powers of primitive roots smaller than $p-1$ generate all DISTINCT elements in $mathbb{Z}_p$? I am aware about what Fermat's little theorem states and about the fact that $mathbb{Z}_p$ has $p-1$ elements in it. Just not sure how to prove that those elements are distinct and I'm having a struggle with finding proper texts about this.






abstract-algebra number-theory elementary-number-theory cryptography







share|cite|improve this question
























  • We define primitive root of $p$ to be a number whose order is $p-1$. Then you should be asking why/how at least one primitive root exists for every prime number..
    – rsadhvika
    Dec 8 at 8:17












  • You can try David Burton's Elementary Number Theory.
    – Brahadeesh
    Dec 8 at 8:25










  • Thank you, I will definetly check this one.
    – R. Jacks
    Dec 8 at 8:37














2












2








2







I'm trying to understand this situation. Why do the powers of primitive roots smaller than $p-1$ generate all DISTINCT elements in $mathbb{Z}_p$? I am aware about what Fermat's little theorem states and about the fact that $mathbb{Z}_p$ has $p-1$ elements in it. Just not sure how to prove that those elements are distinct and I'm having a struggle with finding proper texts about this.






abstract-algebra number-theory elementary-number-theory cryptography







share|cite|improve this question















I'm trying to understand this situation. Why do the powers of primitive roots smaller than $p-1$ generate all DISTINCT elements in $mathbb{Z}_p$? I am aware about what Fermat's little theorem states and about the fact that $mathbb{Z}_p$ has $p-1$ elements in it. Just not sure how to prove that those elements are distinct and I'm having a struggle with finding proper texts about this.






abstract-algebra number-theory elementary-number-theory cryptography




abstract-algebra number-theory elementary-number-theory cryptography






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 8 at 8:25









Brahadeesh

6,11742360




6,11742360










asked Dec 8 at 8:10









R. Jacks

112




112












  • We define primitive root of $p$ to be a number whose order is $p-1$. Then you should be asking why/how at least one primitive root exists for every prime number..
    – rsadhvika
    Dec 8 at 8:17












  • You can try David Burton's Elementary Number Theory.
    – Brahadeesh
    Dec 8 at 8:25










  • Thank you, I will definetly check this one.
    – R. Jacks
    Dec 8 at 8:37


















  • We define primitive root of $p$ to be a number whose order is $p-1$. Then you should be asking why/how at least one primitive root exists for every prime number..
    – rsadhvika
    Dec 8 at 8:17












  • You can try David Burton's Elementary Number Theory.
    – Brahadeesh
    Dec 8 at 8:25










  • Thank you, I will definetly check this one.
    – R. Jacks
    Dec 8 at 8:37
















We define primitive root of $p$ to be a number whose order is $p-1$. Then you should be asking why/how at least one primitive root exists for every prime number..
– rsadhvika
Dec 8 at 8:17






We define primitive root of $p$ to be a number whose order is $p-1$. Then you should be asking why/how at least one primitive root exists for every prime number..
– rsadhvika
Dec 8 at 8:17














You can try David Burton's Elementary Number Theory.
– Brahadeesh
Dec 8 at 8:25




You can try David Burton's Elementary Number Theory.
– Brahadeesh
Dec 8 at 8:25












Thank you, I will definetly check this one.
– R. Jacks
Dec 8 at 8:37




Thank you, I will definetly check this one.
– R. Jacks
Dec 8 at 8:37










2 Answers
2






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oldest

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1














Suppose $r^aequiv r^b pmod p $ for some $1leq a,bleq p-1$ . Since $(r^b,p)=1$,inverse of $r^b$ exists, say, $r^{-b}$. Multiplying the first equation by $r^{-b}$ will give us $r^{a-b}equiv 1 pmod p $. Since $r $ is a primitive root and $1leq a,bleq p-1$, the only possibility is $a=b $. Hence, $r^x pmod p$ produces distinct elements as $x$ varies from $1$ to $p-1$. Since there are $p-1$ distinct elements, these elements corresponds the elements of the set ${1,2,ldots,p-1} $.






share|cite|improve this answer

















  • 1




    Thank you! This was a big help.
    – R. Jacks
    Dec 9 at 11:59



















1














We have a group of order $p-1$ (the non-zero class mod $p$ under multiplication) and by definition a primitive root is an element of order $p-1$, so $x^{p-1} =1$ and $x^n neq 1$ for all $1 le n < p-1$ ($p-1$ is the minimal power that gives $1$). Distinctness is then simpel group theory:



If $x^r = x^s$ for some $1le s < r < p-1$ then $x^rcdot (x^s)^{-1} = x^{(r-s)}=1$ and as $1 le r-s < p-1$ this contradicts the minimality condition of the order. So all $x^r$ are distinct for $1 le r < p-1$.






share|cite|improve this answer





















  • Thank you! Got this one explained and now I can continue with existence.
    – R. Jacks
    Dec 9 at 11:59











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2 Answers
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2 Answers
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active

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active

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1














Suppose $r^aequiv r^b pmod p $ for some $1leq a,bleq p-1$ . Since $(r^b,p)=1$,inverse of $r^b$ exists, say, $r^{-b}$. Multiplying the first equation by $r^{-b}$ will give us $r^{a-b}equiv 1 pmod p $. Since $r $ is a primitive root and $1leq a,bleq p-1$, the only possibility is $a=b $. Hence, $r^x pmod p$ produces distinct elements as $x$ varies from $1$ to $p-1$. Since there are $p-1$ distinct elements, these elements corresponds the elements of the set ${1,2,ldots,p-1} $.






share|cite|improve this answer

















  • 1




    Thank you! This was a big help.
    – R. Jacks
    Dec 9 at 11:59
















1














Suppose $r^aequiv r^b pmod p $ for some $1leq a,bleq p-1$ . Since $(r^b,p)=1$,inverse of $r^b$ exists, say, $r^{-b}$. Multiplying the first equation by $r^{-b}$ will give us $r^{a-b}equiv 1 pmod p $. Since $r $ is a primitive root and $1leq a,bleq p-1$, the only possibility is $a=b $. Hence, $r^x pmod p$ produces distinct elements as $x$ varies from $1$ to $p-1$. Since there are $p-1$ distinct elements, these elements corresponds the elements of the set ${1,2,ldots,p-1} $.






share|cite|improve this answer

















  • 1




    Thank you! This was a big help.
    – R. Jacks
    Dec 9 at 11:59














1












1








1






Suppose $r^aequiv r^b pmod p $ for some $1leq a,bleq p-1$ . Since $(r^b,p)=1$,inverse of $r^b$ exists, say, $r^{-b}$. Multiplying the first equation by $r^{-b}$ will give us $r^{a-b}equiv 1 pmod p $. Since $r $ is a primitive root and $1leq a,bleq p-1$, the only possibility is $a=b $. Hence, $r^x pmod p$ produces distinct elements as $x$ varies from $1$ to $p-1$. Since there are $p-1$ distinct elements, these elements corresponds the elements of the set ${1,2,ldots,p-1} $.






share|cite|improve this answer












Suppose $r^aequiv r^b pmod p $ for some $1leq a,bleq p-1$ . Since $(r^b,p)=1$,inverse of $r^b$ exists, say, $r^{-b}$. Multiplying the first equation by $r^{-b}$ will give us $r^{a-b}equiv 1 pmod p $. Since $r $ is a primitive root and $1leq a,bleq p-1$, the only possibility is $a=b $. Hence, $r^x pmod p$ produces distinct elements as $x$ varies from $1$ to $p-1$. Since there are $p-1$ distinct elements, these elements corresponds the elements of the set ${1,2,ldots,p-1} $.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 at 9:51









Thomas Shelby

1,242116




1,242116








  • 1




    Thank you! This was a big help.
    – R. Jacks
    Dec 9 at 11:59














  • 1




    Thank you! This was a big help.
    – R. Jacks
    Dec 9 at 11:59








1




1




Thank you! This was a big help.
– R. Jacks
Dec 9 at 11:59




Thank you! This was a big help.
– R. Jacks
Dec 9 at 11:59











1














We have a group of order $p-1$ (the non-zero class mod $p$ under multiplication) and by definition a primitive root is an element of order $p-1$, so $x^{p-1} =1$ and $x^n neq 1$ for all $1 le n < p-1$ ($p-1$ is the minimal power that gives $1$). Distinctness is then simpel group theory:



If $x^r = x^s$ for some $1le s < r < p-1$ then $x^rcdot (x^s)^{-1} = x^{(r-s)}=1$ and as $1 le r-s < p-1$ this contradicts the minimality condition of the order. So all $x^r$ are distinct for $1 le r < p-1$.






share|cite|improve this answer





















  • Thank you! Got this one explained and now I can continue with existence.
    – R. Jacks
    Dec 9 at 11:59
















1














We have a group of order $p-1$ (the non-zero class mod $p$ under multiplication) and by definition a primitive root is an element of order $p-1$, so $x^{p-1} =1$ and $x^n neq 1$ for all $1 le n < p-1$ ($p-1$ is the minimal power that gives $1$). Distinctness is then simpel group theory:



If $x^r = x^s$ for some $1le s < r < p-1$ then $x^rcdot (x^s)^{-1} = x^{(r-s)}=1$ and as $1 le r-s < p-1$ this contradicts the minimality condition of the order. So all $x^r$ are distinct for $1 le r < p-1$.






share|cite|improve this answer





















  • Thank you! Got this one explained and now I can continue with existence.
    – R. Jacks
    Dec 9 at 11:59














1












1








1






We have a group of order $p-1$ (the non-zero class mod $p$ under multiplication) and by definition a primitive root is an element of order $p-1$, so $x^{p-1} =1$ and $x^n neq 1$ for all $1 le n < p-1$ ($p-1$ is the minimal power that gives $1$). Distinctness is then simpel group theory:



If $x^r = x^s$ for some $1le s < r < p-1$ then $x^rcdot (x^s)^{-1} = x^{(r-s)}=1$ and as $1 le r-s < p-1$ this contradicts the minimality condition of the order. So all $x^r$ are distinct for $1 le r < p-1$.






share|cite|improve this answer












We have a group of order $p-1$ (the non-zero class mod $p$ under multiplication) and by definition a primitive root is an element of order $p-1$, so $x^{p-1} =1$ and $x^n neq 1$ for all $1 le n < p-1$ ($p-1$ is the minimal power that gives $1$). Distinctness is then simpel group theory:



If $x^r = x^s$ for some $1le s < r < p-1$ then $x^rcdot (x^s)^{-1} = x^{(r-s)}=1$ and as $1 le r-s < p-1$ this contradicts the minimality condition of the order. So all $x^r$ are distinct for $1 le r < p-1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 at 22:44









Henno Brandsma

104k346113




104k346113












  • Thank you! Got this one explained and now I can continue with existence.
    – R. Jacks
    Dec 9 at 11:59


















  • Thank you! Got this one explained and now I can continue with existence.
    – R. Jacks
    Dec 9 at 11:59
















Thank you! Got this one explained and now I can continue with existence.
– R. Jacks
Dec 9 at 11:59




Thank you! Got this one explained and now I can continue with existence.
– R. Jacks
Dec 9 at 11:59


















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