How to verify if g is a generator for p?












3












$begingroup$


For learning purpose, supposed I have a 16-digit prime which is 2685735182215187, how do I verify if g is a generator? (p is supposedly a special kind of prime)






rsa prime-numbers elgamal-encryption







share|improve this question









$endgroup$








  • 1




    $begingroup$
    The special kind of prime that you have is called a safe prime. it's a prime of the form $p = 2q + 1$ where $q$ is also prime (as shown by poncho's answer).
    $endgroup$
    – puzzlepalace
    14 hours ago










  • $begingroup$
    @puzzlepalace sorry, I'm still confused about q. Where do I actually get the q?
    $endgroup$
    – Ken
    13 hours ago










  • $begingroup$
    You can derive $q$ from $p$. In other words, to check if $p$ is a safe prime, you check if $q = frac{p-1}{2}$ is also prime.
    $endgroup$
    – puzzlepalace
    13 hours ago










  • $begingroup$
    @puzzlepalace Thank you for your swift reply. I have computed and checked q=(p-1)/2 and my program returns true (it is indeed a prime). So I'm safe to say that q is also a prime, which means that p is a special kind of prime.
    $endgroup$
    – Ken
    12 hours ago










  • $begingroup$
    @puzzlepalace However, I'm still confused about g. I have computed g^(p-1)/2 mod p and g^p-1/(p-1/2) like what poncho has mentioned. The first output is 1342867591052455, and the second output is 0. I'm a little confused about these numbers, do they mean that g is a generator?
    $endgroup$
    – Ken
    12 hours ago
















3












$begingroup$


For learning purpose, supposed I have a 16-digit prime which is 2685735182215187, how do I verify if g is a generator? (p is supposedly a special kind of prime)






rsa prime-numbers elgamal-encryption







share|improve this question









$endgroup$








  • 1




    $begingroup$
    The special kind of prime that you have is called a safe prime. it's a prime of the form $p = 2q + 1$ where $q$ is also prime (as shown by poncho's answer).
    $endgroup$
    – puzzlepalace
    14 hours ago










  • $begingroup$
    @puzzlepalace sorry, I'm still confused about q. Where do I actually get the q?
    $endgroup$
    – Ken
    13 hours ago










  • $begingroup$
    You can derive $q$ from $p$. In other words, to check if $p$ is a safe prime, you check if $q = frac{p-1}{2}$ is also prime.
    $endgroup$
    – puzzlepalace
    13 hours ago










  • $begingroup$
    @puzzlepalace Thank you for your swift reply. I have computed and checked q=(p-1)/2 and my program returns true (it is indeed a prime). So I'm safe to say that q is also a prime, which means that p is a special kind of prime.
    $endgroup$
    – Ken
    12 hours ago










  • $begingroup$
    @puzzlepalace However, I'm still confused about g. I have computed g^(p-1)/2 mod p and g^p-1/(p-1/2) like what poncho has mentioned. The first output is 1342867591052455, and the second output is 0. I'm a little confused about these numbers, do they mean that g is a generator?
    $endgroup$
    – Ken
    12 hours ago














3












3








3





$begingroup$


For learning purpose, supposed I have a 16-digit prime which is 2685735182215187, how do I verify if g is a generator? (p is supposedly a special kind of prime)






rsa prime-numbers elgamal-encryption







share|improve this question









$endgroup$




For learning purpose, supposed I have a 16-digit prime which is 2685735182215187, how do I verify if g is a generator? (p is supposedly a special kind of prime)






rsa prime-numbers elgamal-encryption




rsa prime-numbers elgamal-encryption






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 15 hours ago









KenKen

312




312








  • 1




    $begingroup$
    The special kind of prime that you have is called a safe prime. it's a prime of the form $p = 2q + 1$ where $q$ is also prime (as shown by poncho's answer).
    $endgroup$
    – puzzlepalace
    14 hours ago










  • $begingroup$
    @puzzlepalace sorry, I'm still confused about q. Where do I actually get the q?
    $endgroup$
    – Ken
    13 hours ago










  • $begingroup$
    You can derive $q$ from $p$. In other words, to check if $p$ is a safe prime, you check if $q = frac{p-1}{2}$ is also prime.
    $endgroup$
    – puzzlepalace
    13 hours ago










  • $begingroup$
    @puzzlepalace Thank you for your swift reply. I have computed and checked q=(p-1)/2 and my program returns true (it is indeed a prime). So I'm safe to say that q is also a prime, which means that p is a special kind of prime.
    $endgroup$
    – Ken
    12 hours ago










  • $begingroup$
    @puzzlepalace However, I'm still confused about g. I have computed g^(p-1)/2 mod p and g^p-1/(p-1/2) like what poncho has mentioned. The first output is 1342867591052455, and the second output is 0. I'm a little confused about these numbers, do they mean that g is a generator?
    $endgroup$
    – Ken
    12 hours ago














  • 1




    $begingroup$
    The special kind of prime that you have is called a safe prime. it's a prime of the form $p = 2q + 1$ where $q$ is also prime (as shown by poncho's answer).
    $endgroup$
    – puzzlepalace
    14 hours ago










  • $begingroup$
    @puzzlepalace sorry, I'm still confused about q. Where do I actually get the q?
    $endgroup$
    – Ken
    13 hours ago










  • $begingroup$
    You can derive $q$ from $p$. In other words, to check if $p$ is a safe prime, you check if $q = frac{p-1}{2}$ is also prime.
    $endgroup$
    – puzzlepalace
    13 hours ago










  • $begingroup$
    @puzzlepalace Thank you for your swift reply. I have computed and checked q=(p-1)/2 and my program returns true (it is indeed a prime). So I'm safe to say that q is also a prime, which means that p is a special kind of prime.
    $endgroup$
    – Ken
    12 hours ago










  • $begingroup$
    @puzzlepalace However, I'm still confused about g. I have computed g^(p-1)/2 mod p and g^p-1/(p-1/2) like what poncho has mentioned. The first output is 1342867591052455, and the second output is 0. I'm a little confused about these numbers, do they mean that g is a generator?
    $endgroup$
    – Ken
    12 hours ago








1




1




$begingroup$
The special kind of prime that you have is called a safe prime. it's a prime of the form $p = 2q + 1$ where $q$ is also prime (as shown by poncho's answer).
$endgroup$
– puzzlepalace
14 hours ago




$begingroup$
The special kind of prime that you have is called a safe prime. it's a prime of the form $p = 2q + 1$ where $q$ is also prime (as shown by poncho's answer).
$endgroup$
– puzzlepalace
14 hours ago












$begingroup$
@puzzlepalace sorry, I'm still confused about q. Where do I actually get the q?
$endgroup$
– Ken
13 hours ago




$begingroup$
@puzzlepalace sorry, I'm still confused about q. Where do I actually get the q?
$endgroup$
– Ken
13 hours ago












$begingroup$
You can derive $q$ from $p$. In other words, to check if $p$ is a safe prime, you check if $q = frac{p-1}{2}$ is also prime.
$endgroup$
– puzzlepalace
13 hours ago




$begingroup$
You can derive $q$ from $p$. In other words, to check if $p$ is a safe prime, you check if $q = frac{p-1}{2}$ is also prime.
$endgroup$
– puzzlepalace
13 hours ago












$begingroup$
@puzzlepalace Thank you for your swift reply. I have computed and checked q=(p-1)/2 and my program returns true (it is indeed a prime). So I'm safe to say that q is also a prime, which means that p is a special kind of prime.
$endgroup$
– Ken
12 hours ago




$begingroup$
@puzzlepalace Thank you for your swift reply. I have computed and checked q=(p-1)/2 and my program returns true (it is indeed a prime). So I'm safe to say that q is also a prime, which means that p is a special kind of prime.
$endgroup$
– Ken
12 hours ago












$begingroup$
@puzzlepalace However, I'm still confused about g. I have computed g^(p-1)/2 mod p and g^p-1/(p-1/2) like what poncho has mentioned. The first output is 1342867591052455, and the second output is 0. I'm a little confused about these numbers, do they mean that g is a generator?
$endgroup$
– Ken
12 hours ago




$begingroup$
@puzzlepalace However, I'm still confused about g. I have computed g^(p-1)/2 mod p and g^p-1/(p-1/2) like what poncho has mentioned. The first output is 1342867591052455, and the second output is 0. I'm a little confused about these numbers, do they mean that g is a generator?
$endgroup$
– Ken
12 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

Steps:




  • Factor $p-1$, that is, find the primes which, multiplied together, produce $p-1$. In your case, $2685735182215186 = 2 times 1342867591107593$


  • For each prime factor $q$ of $p-1$, verify that $g^{(p-1)/q} notequiv 1pmod p$



If every such $q$ verifies (that is, they were all not 1), then $g$ is a generator.






share|improve this answer









$endgroup$













  • $begingroup$
    Hey @poncho thanks. I do not understand "For each prime factor q of p−1, verify that g(p−1)/q≢1(mod p)" Is there anyway you can explain it simpler?
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Compute $g^{2685735182215186/2} bmod p$. Compute $g^{2685735182215186/1342867591107593} bmod p$. If they are both something other than 1, then $g$ is a generator
    $endgroup$
    – poncho
    14 hours ago












  • $begingroup$
    Thank you so much @poncho
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Java's long type has 64 bits; it's not going to be able to store $2^{1342867591107593}$ without wrapping around. You will need to either switch to BigIntegers (in which case you really should use BigInteger::modPow) or implement a modular exponentiation algorithm yourself.
    $endgroup$
    – Ilmari Karonen
    11 hours ago








  • 1




    $begingroup$
    What if factoring $p - 1$ is unfeasible? Is it then impossible to verify or are there other techniques you can apply?
    $endgroup$
    – orlp
    8 hours ago



















3












$begingroup$

In general, proving that $g$ is a primitive root (often called a generator) of a cyclic group is fairly simple. Note this holds true for non prime modulo as well



Step 1:



Verify that $0leqslant g lt p$ and $(g,p)=1$



In other words, verify that $g$ is less than p but greater than or equal to 0, and that $g$ and $p$ are coprime.



Where $g$ is the element of the group in question and p is the modulus being used (or: $mathbb{Z}_p$).



Step 2:



Calculate $phi(p)$ where $phi$ is the Totient Function. If it happens that $p$ is prime, $phi(p)=p-1$



Then break $phi(p)$ into it's prime factors such that $phi(p)=prodlimits_{i}q_i^{r_i}$ Where each $q_i$ is a prime factor and $r_i$ is the power that prime factor is raised to.



Verify that $g^{phi(p)/q_i}notequiv 1 (mod p)$ $forall q_i$



Ignore the power $r_i$ for this calculation.



Assuming these conditions are met, $g$ is a generator of $mathbb{Z}_p$.






share|improve this answer








New contributor




TryingToPassCollege is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Hi @TryingToPassCollege, thank you so much. However, could you give an example? For learning purpose, for example, p = 2685735182104907 and g = 2.. I understand from Step 1 that from the looks of my p and g, it is definitely between 0 and p, and they are definitely coprime because I made a primality check on Java, and p is a prime. As such, g = 2, is a coprime as well. From step 2 onwards, I'm a little confused because tbh I don't understand most of the symbols. I feel like I'm lacking a lot of mathematics experience.. So sorry for all the trouble, as I don't have anyone else to turn to.
    $endgroup$
    – Ken
    2 hours ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Steps:




  • Factor $p-1$, that is, find the primes which, multiplied together, produce $p-1$. In your case, $2685735182215186 = 2 times 1342867591107593$


  • For each prime factor $q$ of $p-1$, verify that $g^{(p-1)/q} notequiv 1pmod p$



If every such $q$ verifies (that is, they were all not 1), then $g$ is a generator.






share|improve this answer









$endgroup$













  • $begingroup$
    Hey @poncho thanks. I do not understand "For each prime factor q of p−1, verify that g(p−1)/q≢1(mod p)" Is there anyway you can explain it simpler?
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Compute $g^{2685735182215186/2} bmod p$. Compute $g^{2685735182215186/1342867591107593} bmod p$. If they are both something other than 1, then $g$ is a generator
    $endgroup$
    – poncho
    14 hours ago












  • $begingroup$
    Thank you so much @poncho
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Java's long type has 64 bits; it's not going to be able to store $2^{1342867591107593}$ without wrapping around. You will need to either switch to BigIntegers (in which case you really should use BigInteger::modPow) or implement a modular exponentiation algorithm yourself.
    $endgroup$
    – Ilmari Karonen
    11 hours ago








  • 1




    $begingroup$
    What if factoring $p - 1$ is unfeasible? Is it then impossible to verify or are there other techniques you can apply?
    $endgroup$
    – orlp
    8 hours ago
















5












$begingroup$

Steps:




  • Factor $p-1$, that is, find the primes which, multiplied together, produce $p-1$. In your case, $2685735182215186 = 2 times 1342867591107593$


  • For each prime factor $q$ of $p-1$, verify that $g^{(p-1)/q} notequiv 1pmod p$



If every such $q$ verifies (that is, they were all not 1), then $g$ is a generator.






share|improve this answer









$endgroup$













  • $begingroup$
    Hey @poncho thanks. I do not understand "For each prime factor q of p−1, verify that g(p−1)/q≢1(mod p)" Is there anyway you can explain it simpler?
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Compute $g^{2685735182215186/2} bmod p$. Compute $g^{2685735182215186/1342867591107593} bmod p$. If they are both something other than 1, then $g$ is a generator
    $endgroup$
    – poncho
    14 hours ago












  • $begingroup$
    Thank you so much @poncho
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Java's long type has 64 bits; it's not going to be able to store $2^{1342867591107593}$ without wrapping around. You will need to either switch to BigIntegers (in which case you really should use BigInteger::modPow) or implement a modular exponentiation algorithm yourself.
    $endgroup$
    – Ilmari Karonen
    11 hours ago








  • 1




    $begingroup$
    What if factoring $p - 1$ is unfeasible? Is it then impossible to verify or are there other techniques you can apply?
    $endgroup$
    – orlp
    8 hours ago














5












5








5





$begingroup$

Steps:




  • Factor $p-1$, that is, find the primes which, multiplied together, produce $p-1$. In your case, $2685735182215186 = 2 times 1342867591107593$


  • For each prime factor $q$ of $p-1$, verify that $g^{(p-1)/q} notequiv 1pmod p$



If every such $q$ verifies (that is, they were all not 1), then $g$ is a generator.






share|improve this answer









$endgroup$



Steps:




  • Factor $p-1$, that is, find the primes which, multiplied together, produce $p-1$. In your case, $2685735182215186 = 2 times 1342867591107593$


  • For each prime factor $q$ of $p-1$, verify that $g^{(p-1)/q} notequiv 1pmod p$



If every such $q$ verifies (that is, they were all not 1), then $g$ is a generator.







share|improve this answer












share|improve this answer



share|improve this answer










answered 15 hours ago









ponchoponcho

93.4k2146242




93.4k2146242












  • $begingroup$
    Hey @poncho thanks. I do not understand "For each prime factor q of p−1, verify that g(p−1)/q≢1(mod p)" Is there anyway you can explain it simpler?
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Compute $g^{2685735182215186/2} bmod p$. Compute $g^{2685735182215186/1342867591107593} bmod p$. If they are both something other than 1, then $g$ is a generator
    $endgroup$
    – poncho
    14 hours ago












  • $begingroup$
    Thank you so much @poncho
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Java's long type has 64 bits; it's not going to be able to store $2^{1342867591107593}$ without wrapping around. You will need to either switch to BigIntegers (in which case you really should use BigInteger::modPow) or implement a modular exponentiation algorithm yourself.
    $endgroup$
    – Ilmari Karonen
    11 hours ago








  • 1




    $begingroup$
    What if factoring $p - 1$ is unfeasible? Is it then impossible to verify or are there other techniques you can apply?
    $endgroup$
    – orlp
    8 hours ago


















  • $begingroup$
    Hey @poncho thanks. I do not understand "For each prime factor q of p−1, verify that g(p−1)/q≢1(mod p)" Is there anyway you can explain it simpler?
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Compute $g^{2685735182215186/2} bmod p$. Compute $g^{2685735182215186/1342867591107593} bmod p$. If they are both something other than 1, then $g$ is a generator
    $endgroup$
    – poncho
    14 hours ago












  • $begingroup$
    Thank you so much @poncho
    $endgroup$
    – Ken
    14 hours ago










  • $begingroup$
    @Ken: Java's long type has 64 bits; it's not going to be able to store $2^{1342867591107593}$ without wrapping around. You will need to either switch to BigIntegers (in which case you really should use BigInteger::modPow) or implement a modular exponentiation algorithm yourself.
    $endgroup$
    – Ilmari Karonen
    11 hours ago








  • 1




    $begingroup$
    What if factoring $p - 1$ is unfeasible? Is it then impossible to verify or are there other techniques you can apply?
    $endgroup$
    – orlp
    8 hours ago
















$begingroup$
Hey @poncho thanks. I do not understand "For each prime factor q of p−1, verify that g(p−1)/q≢1(mod p)" Is there anyway you can explain it simpler?
$endgroup$
– Ken
14 hours ago




$begingroup$
Hey @poncho thanks. I do not understand "For each prime factor q of p−1, verify that g(p−1)/q≢1(mod p)" Is there anyway you can explain it simpler?
$endgroup$
– Ken
14 hours ago












$begingroup$
@Ken: Compute $g^{2685735182215186/2} bmod p$. Compute $g^{2685735182215186/1342867591107593} bmod p$. If they are both something other than 1, then $g$ is a generator
$endgroup$
– poncho
14 hours ago






$begingroup$
@Ken: Compute $g^{2685735182215186/2} bmod p$. Compute $g^{2685735182215186/1342867591107593} bmod p$. If they are both something other than 1, then $g$ is a generator
$endgroup$
– poncho
14 hours ago














$begingroup$
Thank you so much @poncho
$endgroup$
– Ken
14 hours ago




$begingroup$
Thank you so much @poncho
$endgroup$
– Ken
14 hours ago












$begingroup$
@Ken: Java's long type has 64 bits; it's not going to be able to store $2^{1342867591107593}$ without wrapping around. You will need to either switch to BigIntegers (in which case you really should use BigInteger::modPow) or implement a modular exponentiation algorithm yourself.
$endgroup$
– Ilmari Karonen
11 hours ago






$begingroup$
@Ken: Java's long type has 64 bits; it's not going to be able to store $2^{1342867591107593}$ without wrapping around. You will need to either switch to BigIntegers (in which case you really should use BigInteger::modPow) or implement a modular exponentiation algorithm yourself.
$endgroup$
– Ilmari Karonen
11 hours ago






1




1




$begingroup$
What if factoring $p - 1$ is unfeasible? Is it then impossible to verify or are there other techniques you can apply?
$endgroup$
– orlp
8 hours ago




$begingroup$
What if factoring $p - 1$ is unfeasible? Is it then impossible to verify or are there other techniques you can apply?
$endgroup$
– orlp
8 hours ago











3












$begingroup$

In general, proving that $g$ is a primitive root (often called a generator) of a cyclic group is fairly simple. Note this holds true for non prime modulo as well



Step 1:



Verify that $0leqslant g lt p$ and $(g,p)=1$



In other words, verify that $g$ is less than p but greater than or equal to 0, and that $g$ and $p$ are coprime.



Where $g$ is the element of the group in question and p is the modulus being used (or: $mathbb{Z}_p$).



Step 2:



Calculate $phi(p)$ where $phi$ is the Totient Function. If it happens that $p$ is prime, $phi(p)=p-1$



Then break $phi(p)$ into it's prime factors such that $phi(p)=prodlimits_{i}q_i^{r_i}$ Where each $q_i$ is a prime factor and $r_i$ is the power that prime factor is raised to.



Verify that $g^{phi(p)/q_i}notequiv 1 (mod p)$ $forall q_i$



Ignore the power $r_i$ for this calculation.



Assuming these conditions are met, $g$ is a generator of $mathbb{Z}_p$.






share|improve this answer








New contributor




TryingToPassCollege is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Hi @TryingToPassCollege, thank you so much. However, could you give an example? For learning purpose, for example, p = 2685735182104907 and g = 2.. I understand from Step 1 that from the looks of my p and g, it is definitely between 0 and p, and they are definitely coprime because I made a primality check on Java, and p is a prime. As such, g = 2, is a coprime as well. From step 2 onwards, I'm a little confused because tbh I don't understand most of the symbols. I feel like I'm lacking a lot of mathematics experience.. So sorry for all the trouble, as I don't have anyone else to turn to.
    $endgroup$
    – Ken
    2 hours ago


















3












$begingroup$

In general, proving that $g$ is a primitive root (often called a generator) of a cyclic group is fairly simple. Note this holds true for non prime modulo as well



Step 1:



Verify that $0leqslant g lt p$ and $(g,p)=1$



In other words, verify that $g$ is less than p but greater than or equal to 0, and that $g$ and $p$ are coprime.



Where $g$ is the element of the group in question and p is the modulus being used (or: $mathbb{Z}_p$).



Step 2:



Calculate $phi(p)$ where $phi$ is the Totient Function. If it happens that $p$ is prime, $phi(p)=p-1$



Then break $phi(p)$ into it's prime factors such that $phi(p)=prodlimits_{i}q_i^{r_i}$ Where each $q_i$ is a prime factor and $r_i$ is the power that prime factor is raised to.



Verify that $g^{phi(p)/q_i}notequiv 1 (mod p)$ $forall q_i$



Ignore the power $r_i$ for this calculation.



Assuming these conditions are met, $g$ is a generator of $mathbb{Z}_p$.






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New contributor




TryingToPassCollege is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$













  • $begingroup$
    Hi @TryingToPassCollege, thank you so much. However, could you give an example? For learning purpose, for example, p = 2685735182104907 and g = 2.. I understand from Step 1 that from the looks of my p and g, it is definitely between 0 and p, and they are definitely coprime because I made a primality check on Java, and p is a prime. As such, g = 2, is a coprime as well. From step 2 onwards, I'm a little confused because tbh I don't understand most of the symbols. I feel like I'm lacking a lot of mathematics experience.. So sorry for all the trouble, as I don't have anyone else to turn to.
    $endgroup$
    – Ken
    2 hours ago
















3












3








3





$begingroup$

In general, proving that $g$ is a primitive root (often called a generator) of a cyclic group is fairly simple. Note this holds true for non prime modulo as well



Step 1:



Verify that $0leqslant g lt p$ and $(g,p)=1$



In other words, verify that $g$ is less than p but greater than or equal to 0, and that $g$ and $p$ are coprime.



Where $g$ is the element of the group in question and p is the modulus being used (or: $mathbb{Z}_p$).



Step 2:



Calculate $phi(p)$ where $phi$ is the Totient Function. If it happens that $p$ is prime, $phi(p)=p-1$



Then break $phi(p)$ into it's prime factors such that $phi(p)=prodlimits_{i}q_i^{r_i}$ Where each $q_i$ is a prime factor and $r_i$ is the power that prime factor is raised to.



Verify that $g^{phi(p)/q_i}notequiv 1 (mod p)$ $forall q_i$



Ignore the power $r_i$ for this calculation.



Assuming these conditions are met, $g$ is a generator of $mathbb{Z}_p$.






share|improve this answer








New contributor




TryingToPassCollege is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



In general, proving that $g$ is a primitive root (often called a generator) of a cyclic group is fairly simple. Note this holds true for non prime modulo as well



Step 1:



Verify that $0leqslant g lt p$ and $(g,p)=1$



In other words, verify that $g$ is less than p but greater than or equal to 0, and that $g$ and $p$ are coprime.



Where $g$ is the element of the group in question and p is the modulus being used (or: $mathbb{Z}_p$).



Step 2:



Calculate $phi(p)$ where $phi$ is the Totient Function. If it happens that $p$ is prime, $phi(p)=p-1$



Then break $phi(p)$ into it's prime factors such that $phi(p)=prodlimits_{i}q_i^{r_i}$ Where each $q_i$ is a prime factor and $r_i$ is the power that prime factor is raised to.



Verify that $g^{phi(p)/q_i}notequiv 1 (mod p)$ $forall q_i$



Ignore the power $r_i$ for this calculation.



Assuming these conditions are met, $g$ is a generator of $mathbb{Z}_p$.







share|improve this answer








New contributor




TryingToPassCollege is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this answer



share|improve this answer






New contributor




TryingToPassCollege is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 8 hours ago









TryingToPassCollegeTryingToPassCollege

411




411




New contributor




TryingToPassCollege is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





TryingToPassCollege is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






TryingToPassCollege is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Hi @TryingToPassCollege, thank you so much. However, could you give an example? For learning purpose, for example, p = 2685735182104907 and g = 2.. I understand from Step 1 that from the looks of my p and g, it is definitely between 0 and p, and they are definitely coprime because I made a primality check on Java, and p is a prime. As such, g = 2, is a coprime as well. From step 2 onwards, I'm a little confused because tbh I don't understand most of the symbols. I feel like I'm lacking a lot of mathematics experience.. So sorry for all the trouble, as I don't have anyone else to turn to.
    $endgroup$
    – Ken
    2 hours ago




















  • $begingroup$
    Hi @TryingToPassCollege, thank you so much. However, could you give an example? For learning purpose, for example, p = 2685735182104907 and g = 2.. I understand from Step 1 that from the looks of my p and g, it is definitely between 0 and p, and they are definitely coprime because I made a primality check on Java, and p is a prime. As such, g = 2, is a coprime as well. From step 2 onwards, I'm a little confused because tbh I don't understand most of the symbols. I feel like I'm lacking a lot of mathematics experience.. So sorry for all the trouble, as I don't have anyone else to turn to.
    $endgroup$
    – Ken
    2 hours ago


















$begingroup$
Hi @TryingToPassCollege, thank you so much. However, could you give an example? For learning purpose, for example, p = 2685735182104907 and g = 2.. I understand from Step 1 that from the looks of my p and g, it is definitely between 0 and p, and they are definitely coprime because I made a primality check on Java, and p is a prime. As such, g = 2, is a coprime as well. From step 2 onwards, I'm a little confused because tbh I don't understand most of the symbols. I feel like I'm lacking a lot of mathematics experience.. So sorry for all the trouble, as I don't have anyone else to turn to.
$endgroup$
– Ken
2 hours ago






$begingroup$
Hi @TryingToPassCollege, thank you so much. However, could you give an example? For learning purpose, for example, p = 2685735182104907 and g = 2.. I understand from Step 1 that from the looks of my p and g, it is definitely between 0 and p, and they are definitely coprime because I made a primality check on Java, and p is a prime. As such, g = 2, is a coprime as well. From step 2 onwards, I'm a little confused because tbh I don't understand most of the symbols. I feel like I'm lacking a lot of mathematics experience.. So sorry for all the trouble, as I don't have anyone else to turn to.
$endgroup$
– Ken
2 hours ago




















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