Does classifying an integer as a discrete log require it be part of a multiplicative group? Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Discrete log problem, when we have many examplesFinding where I am in a linear recurrence relationA discrete-log-like problem, with matrices: given $A^k x$, find $k$iterated discrete log problemWhy can ECC key sizes be smaller than RSA keys for similar security?Is the reverse of the “discrete logarithm problem” equally dificult?How to construct a hash function into a cyclic group such that its discrete log is intractable?The computational complexity of discrete logSolving the discrete logarithm problem for a weak groupSolving discrete log in partially known group
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Does classifying an integer as a discrete log require it be part of a multiplicative group?
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Does classifying an integer as a discrete log require it be part of a multiplicative group?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Discrete log problem, when we have many examplesFinding where I am in a linear recurrence relationA discrete-log-like problem, with matrices: given $A^k x$, find $k$iterated discrete log problemWhy can ECC key sizes be smaller than RSA keys for similar security?Is the reverse of the “discrete logarithm problem” equally dificult?How to construct a hash function into a cyclic group such that its discrete log is intractable?The computational complexity of discrete logSolving the discrete logarithm problem for a weak groupSolving discrete log in partially known group
$begingroup$
This question is not a question about the discrete log problem, the generalized discrete log problem, or an additive group.
The confusion is whether any integer can be considered a discrete log or whether a discrete log has as a precondition, that it be part of a multiplicative group. This wikipedia would seem to indicate that the answer is yes.
For example 0 doesn't have a multiplicative inverse and is therefore not part of a multiplicative group.
discrete-logarithm terminology
asked 4 hours ago
JohnGaltJohnGalt
28528
$endgroup$
|
show 2 more comments
$begingroup$
This question is not a question about the discrete log problem, the generalized discrete log problem, or an additive group.
The confusion is whether any integer can be considered a discrete log or whether a discrete log has as a precondition, that it be part of a multiplicative group. This wikipedia would seem to indicate that the answer is yes.
For example 0 doesn't have a multiplicative inverse and is therefore not part of a multiplicative group.
discrete-logarithm terminology
asked 4 hours ago
JohnGaltJohnGalt
28528
$endgroup$
$begingroup$
@kelalaka Would you mind expanding upon "The discrete log is defined according to a base as the logarithm."
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
@kelalaka also if "0 is not a part of the multiplicative group" does that mean that not all integers are part of a discrete log?
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
Whomever down voted my question, I respect the decision, however, it would be helpful if you commented as to why you down voted it.
$endgroup$
– JohnGalt
3 hours ago
$begingroup$
Your question currently has no downvotes. That said, I'm tempted to vote to close it as unclear, since it seems to be based on some kind of a confusion of terminology, and it's literally not clear to me what you're trying to ask. The two answers already below are both correct and well written, but they answer completely different questions.
$endgroup$
– Ilmari Karonen
45 mins ago
$begingroup$
(In particular, every integer is, trivially, both a part of infinitely many different multiplicative groups and the discrete logarithm of infinitely many elements of some multiplicative group with respect to some base. And these two facts have nothing to do with each other.)
$endgroup$
– Ilmari Karonen
37 mins ago
|
show 2 more comments
$begingroup$
This question is not a question about the discrete log problem, the generalized discrete log problem, or an additive group.
The confusion is whether any integer can be considered a discrete log or whether a discrete log has as a precondition, that it be part of a multiplicative group. This wikipedia would seem to indicate that the answer is yes.
For example 0 doesn't have a multiplicative inverse and is therefore not part of a multiplicative group.
discrete-logarithm terminology
asked 4 hours ago
JohnGaltJohnGalt
28528
$endgroup$
This question is not a question about the discrete log problem, the generalized discrete log problem, or an additive group.
The confusion is whether any integer can be considered a discrete log or whether a discrete log has as a precondition, that it be part of a multiplicative group. This wikipedia would seem to indicate that the answer is yes.
For example 0 doesn't have a multiplicative inverse and is therefore not part of a multiplicative group.
discrete-logarithm terminology
discrete-logarithm terminology
asked 4 hours ago
JohnGaltJohnGalt
28528
asked 4 hours ago
JohnGaltJohnGalt
28528
asked 4 hours ago
JohnGaltJohnGalt
28528
asked 4 hours ago
JohnGaltJohnGalt
28528
asked 4 hours ago
JohnGaltJohnGalt
28528
28528
$begingroup$
@kelalaka Would you mind expanding upon "The discrete log is defined according to a base as the logarithm."
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
@kelalaka also if "0 is not a part of the multiplicative group" does that mean that not all integers are part of a discrete log?
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
Whomever down voted my question, I respect the decision, however, it would be helpful if you commented as to why you down voted it.
$endgroup$
– JohnGalt
3 hours ago
$begingroup$
Your question currently has no downvotes. That said, I'm tempted to vote to close it as unclear, since it seems to be based on some kind of a confusion of terminology, and it's literally not clear to me what you're trying to ask. The two answers already below are both correct and well written, but they answer completely different questions.
$endgroup$
– Ilmari Karonen
45 mins ago
$begingroup$
(In particular, every integer is, trivially, both a part of infinitely many different multiplicative groups and the discrete logarithm of infinitely many elements of some multiplicative group with respect to some base. And these two facts have nothing to do with each other.)
$endgroup$
– Ilmari Karonen
37 mins ago
|
show 2 more comments
$begingroup$
@kelalaka Would you mind expanding upon "The discrete log is defined according to a base as the logarithm."
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
@kelalaka also if "0 is not a part of the multiplicative group" does that mean that not all integers are part of a discrete log?
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
Whomever down voted my question, I respect the decision, however, it would be helpful if you commented as to why you down voted it.
$endgroup$
– JohnGalt
3 hours ago
$begingroup$
Your question currently has no downvotes. That said, I'm tempted to vote to close it as unclear, since it seems to be based on some kind of a confusion of terminology, and it's literally not clear to me what you're trying to ask. The two answers already below are both correct and well written, but they answer completely different questions.
$endgroup$
– Ilmari Karonen
45 mins ago
$begingroup$
(In particular, every integer is, trivially, both a part of infinitely many different multiplicative groups and the discrete logarithm of infinitely many elements of some multiplicative group with respect to some base. And these two facts have nothing to do with each other.)
$endgroup$
– Ilmari Karonen
37 mins ago
$begingroup$
@kelalaka Would you mind expanding upon "The discrete log is defined according to a base as the logarithm."
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
@kelalaka Would you mind expanding upon "The discrete log is defined according to a base as the logarithm."
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
@kelalaka also if "0 is not a part of the multiplicative group" does that mean that not all integers are part of a discrete log?
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
@kelalaka also if "0 is not a part of the multiplicative group" does that mean that not all integers are part of a discrete log?
$endgroup$
– JohnGalt
4 hours ago
$begingroup$
Whomever down voted my question, I respect the decision, however, it would be helpful if you commented as to why you down voted it.
$endgroup$
– JohnGalt
3 hours ago
$begingroup$
Whomever down voted my question, I respect the decision, however, it would be helpful if you commented as to why you down voted it.
$endgroup$
– JohnGalt
3 hours ago
$begingroup$
Your question currently has no downvotes. That said, I'm tempted to vote to close it as unclear, since it seems to be based on some kind of a confusion of terminology, and it's literally not clear to me what you're trying to ask. The two answers already below are both correct and well written, but they answer completely different questions.
$endgroup$
– Ilmari Karonen
45 mins ago
$begingroup$
Your question currently has no downvotes. That said, I'm tempted to vote to close it as unclear, since it seems to be based on some kind of a confusion of terminology, and it's literally not clear to me what you're trying to ask. The two answers already below are both correct and well written, but they answer completely different questions.
$endgroup$
– Ilmari Karonen
45 mins ago
$begingroup$
(In particular, every integer is, trivially, both a part of infinitely many different multiplicative groups and the discrete logarithm of infinitely many elements of some multiplicative group with respect to some base. And these two facts have nothing to do with each other.)
$endgroup$
– Ilmari Karonen
37 mins ago
$begingroup$
(In particular, every integer is, trivially, both a part of infinitely many different multiplicative groups and the discrete logarithm of infinitely many elements of some multiplicative group with respect to some base. And these two facts have nothing to do with each other.)
$endgroup$
– Ilmari Karonen
37 mins ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The discrete logarithm $log_b a$ is an integer $x$ such that $b^x = a$. Similarly to the logarithms, we need a base, here $b$.
If the base is a generator of the group $g$ then any element of the group can be written as a power of the $g$ for some $k$, $y = g^k$. Therefore, the discrete log of $y$ according to base $g$ is $k$.
Take a generator $g$ of a multiplicative group $G$ with order $n$, and then take $g'=g^k$ where $gcd(k,n) neq 1$. Now the $g'$ will generate a subgroup $G'leqslant G$, not the full group. Then any element of the full group $ a in G$ and $a notin G'$ has not discrete logarithm according to base $g'$, even it is not a member of the subgroup.
When we consider the non-zero elements of a field $Fbackslash0$ they are forming a cyclic group under multiplication. For a proof see the Theorem 1.
answered 3 hours ago
kelalakakelalaka
8,86532351
$endgroup$
$begingroup$
It is very possible for $g^2$ to be a generator of $G$; in fact it will be one if and only if the order of $G$ is odd.
$endgroup$
– fkraiem
2 hours ago
$begingroup$
@fkraiem updated to guarantee that $g^k$ is generates a subgroup.
$endgroup$
– kelalaka
2 hours ago
add a comment |
$begingroup$
Of course any integer can be a discrete logarithm: in a group $G$ with generator $g$, any integer $x$ is a discrete logarithm of the group element $g^x$.
Another convenient way to consider the set of discrete logarithms is as the ring $mathbf Z/nmathbf Z$, where $n$ is the order of $G$, which makes sense because $g^x = g^x bmod n$ for all $x$. This is especially convenient when $n$ is prime because then the discrete logarithms form a field.
Either way (unless the group is trivial) the discrete logarithms form a non-trivial ring with unity, which is not a group for multiplication.
answered 3 hours ago
fkraiemfkraiem
6,80021732
$endgroup$
$begingroup$
what if n is a square? If n = k^2, then k is not a discrete log mod n.
$endgroup$
– grovkin
2 hours ago
$begingroup$
@grovkin Why not? $k$ is a discrete log of $g^k$. Are you confusing it with quadratic residue?
$endgroup$
– fkraiem
2 hours ago
$begingroup$
I was just looking for an example of a group element which would not generate anything but itself. I guess copies of Z/2Z is what I should have gone with. But it's not an interesting example. What about Z/nZ, where n = p^l? What is the discrete log of 1+p^l-p^(l-1)? p is a prime, of course.
$endgroup$
– grovkin
1 hour ago
$begingroup$
@grovkin To talk about discrete logs, you need a cyclic group and a generator. If your group is Z/nZ (additive) with generator 1, the discrete log of k is k.
$endgroup$
– fkraiem
1 hour ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The discrete logarithm $log_b a$ is an integer $x$ such that $b^x = a$. Similarly to the logarithms, we need a base, here $b$.
If the base is a generator of the group $g$ then any element of the group can be written as a power of the $g$ for some $k$, $y = g^k$. Therefore, the discrete log of $y$ according to base $g$ is $k$.
Take a generator $g$ of a multiplicative group $G$ with order $n$, and then take $g'=g^k$ where $gcd(k,n) neq 1$. Now the $g'$ will generate a subgroup $G'leqslant G$, not the full group. Then any element of the full group $ a in G$ and $a notin G'$ has not discrete logarithm according to base $g'$, even it is not a member of the subgroup.
When we consider the non-zero elements of a field $Fbackslash0$ they are forming a cyclic group under multiplication. For a proof see the Theorem 1.
answered 3 hours ago
kelalakakelalaka
8,86532351
$endgroup$
$begingroup$
It is very possible for $g^2$ to be a generator of $G$; in fact it will be one if and only if the order of $G$ is odd.
$endgroup$
– fkraiem
2 hours ago
$begingroup$
@fkraiem updated to guarantee that $g^k$ is generates a subgroup.
$endgroup$
– kelalaka
2 hours ago
add a comment |
$begingroup$
The discrete logarithm $log_b a$ is an integer $x$ such that $b^x = a$. Similarly to the logarithms, we need a base, here $b$.
If the base is a generator of the group $g$ then any element of the group can be written as a power of the $g$ for some $k$, $y = g^k$. Therefore, the discrete log of $y$ according to base $g$ is $k$.
Take a generator $g$ of a multiplicative group $G$ with order $n$, and then take $g'=g^k$ where $gcd(k,n) neq 1$. Now the $g'$ will generate a subgroup $G'leqslant G$, not the full group. Then any element of the full group $ a in G$ and $a notin G'$ has not discrete logarithm according to base $g'$, even it is not a member of the subgroup.
When we consider the non-zero elements of a field $Fbackslash0$ they are forming a cyclic group under multiplication. For a proof see the Theorem 1.
answered 3 hours ago
kelalakakelalaka
8,86532351
$endgroup$
$begingroup$
It is very possible for $g^2$ to be a generator of $G$; in fact it will be one if and only if the order of $G$ is odd.
$endgroup$
– fkraiem
2 hours ago
$begingroup$
@fkraiem updated to guarantee that $g^k$ is generates a subgroup.
$endgroup$
– kelalaka
2 hours ago
add a comment |
$begingroup$
The discrete logarithm $log_b a$ is an integer $x$ such that $b^x = a$. Similarly to the logarithms, we need a base, here $b$.
If the base is a generator of the group $g$ then any element of the group can be written as a power of the $g$ for some $k$, $y = g^k$. Therefore, the discrete log of $y$ according to base $g$ is $k$.
Take a generator $g$ of a multiplicative group $G$ with order $n$, and then take $g'=g^k$ where $gcd(k,n) neq 1$. Now the $g'$ will generate a subgroup $G'leqslant G$, not the full group. Then any element of the full group $ a in G$ and $a notin G'$ has not discrete logarithm according to base $g'$, even it is not a member of the subgroup.
When we consider the non-zero elements of a field $Fbackslash0$ they are forming a cyclic group under multiplication. For a proof see the Theorem 1.
answered 3 hours ago
kelalakakelalaka
8,86532351
$endgroup$
The discrete logarithm $log_b a$ is an integer $x$ such that $b^x = a$. Similarly to the logarithms, we need a base, here $b$.
If the base is a generator of the group $g$ then any element of the group can be written as a power of the $g$ for some $k$, $y = g^k$. Therefore, the discrete log of $y$ according to base $g$ is $k$.
Take a generator $g$ of a multiplicative group $G$ with order $n$, and then take $g'=g^k$ where $gcd(k,n) neq 1$. Now the $g'$ will generate a subgroup $G'leqslant G$, not the full group. Then any element of the full group $ a in G$ and $a notin G'$ has not discrete logarithm according to base $g'$, even it is not a member of the subgroup.
When we consider the non-zero elements of a field $Fbackslash0$ they are forming a cyclic group under multiplication. For a proof see the Theorem 1.
answered 3 hours ago
kelalakakelalaka
8,86532351
edited 2 hours ago
answered 3 hours ago
kelalakakelalaka
8,86532351
answered 3 hours ago
kelalakakelalaka
8,86532351
answered 3 hours ago
kelalakakelalaka
8,86532351
8,86532351
$begingroup$
It is very possible for $g^2$ to be a generator of $G$; in fact it will be one if and only if the order of $G$ is odd.
$endgroup$
– fkraiem
2 hours ago
$begingroup$
@fkraiem updated to guarantee that $g^k$ is generates a subgroup.
$endgroup$
– kelalaka
2 hours ago
add a comment |
$begingroup$
It is very possible for $g^2$ to be a generator of $G$; in fact it will be one if and only if the order of $G$ is odd.
$endgroup$
– fkraiem
2 hours ago
$begingroup$
@fkraiem updated to guarantee that $g^k$ is generates a subgroup.
$endgroup$
– kelalaka
2 hours ago
$begingroup$
It is very possible for $g^2$ to be a generator of $G$; in fact it will be one if and only if the order of $G$ is odd.
$endgroup$
– fkraiem
2 hours ago
$begingroup$
It is very possible for $g^2$ to be a generator of $G$; in fact it will be one if and only if the order of $G$ is odd.
$endgroup$
– fkraiem
2 hours ago
$begingroup$
@fkraiem updated to guarantee that $g^k$ is generates a subgroup.
$endgroup$
– kelalaka
2 hours ago
$begingroup$
@fkraiem updated to guarantee that $g^k$ is generates a subgroup.
$endgroup$
– kelalaka
2 hours ago
add a comment |
$begingroup$
Of course any integer can be a discrete logarithm: in a group $G$ with generator $g$, any integer $x$ is a discrete logarithm of the group element $g^x$.
Another convenient way to consider the set of discrete logarithms is as the ring $mathbf Z/nmathbf Z$, where $n$ is the order of $G$, which makes sense because $g^x = g^x bmod n$ for all $x$. This is especially convenient when $n$ is prime because then the discrete logarithms form a field.
Either way (unless the group is trivial) the discrete logarithms form a non-trivial ring with unity, which is not a group for multiplication.
answered 3 hours ago
fkraiemfkraiem
6,80021732
$endgroup$
$begingroup$
what if n is a square? If n = k^2, then k is not a discrete log mod n.
$endgroup$
– grovkin
2 hours ago
$begingroup$
@grovkin Why not? $k$ is a discrete log of $g^k$. Are you confusing it with quadratic residue?
$endgroup$
– fkraiem
2 hours ago
$begingroup$
I was just looking for an example of a group element which would not generate anything but itself. I guess copies of Z/2Z is what I should have gone with. But it's not an interesting example. What about Z/nZ, where n = p^l? What is the discrete log of 1+p^l-p^(l-1)? p is a prime, of course.
$endgroup$
– grovkin
1 hour ago
$begingroup$
@grovkin To talk about discrete logs, you need a cyclic group and a generator. If your group is Z/nZ (additive) with generator 1, the discrete log of k is k.
$endgroup$
– fkraiem
1 hour ago
add a comment |
$begingroup$
Of course any integer can be a discrete logarithm: in a group $G$ with generator $g$, any integer $x$ is a discrete logarithm of the group element $g^x$.
Another convenient way to consider the set of discrete logarithms is as the ring $mathbf Z/nmathbf Z$, where $n$ is the order of $G$, which makes sense because $g^x = g^x bmod n$ for all $x$. This is especially convenient when $n$ is prime because then the discrete logarithms form a field.
Either way (unless the group is trivial) the discrete logarithms form a non-trivial ring with unity, which is not a group for multiplication.
answered 3 hours ago
fkraiemfkraiem
6,80021732
$endgroup$
$begingroup$
what if n is a square? If n = k^2, then k is not a discrete log mod n.
$endgroup$
– grovkin
2 hours ago
$begingroup$
@grovkin Why not? $k$ is a discrete log of $g^k$. Are you confusing it with quadratic residue?
$endgroup$
– fkraiem
2 hours ago
$begingroup$
I was just looking for an example of a group element which would not generate anything but itself. I guess copies of Z/2Z is what I should have gone with. But it's not an interesting example. What about Z/nZ, where n = p^l? What is the discrete log of 1+p^l-p^(l-1)? p is a prime, of course.
$endgroup$
– grovkin
1 hour ago
$begingroup$
@grovkin To talk about discrete logs, you need a cyclic group and a generator. If your group is Z/nZ (additive) with generator 1, the discrete log of k is k.
$endgroup$
– fkraiem
1 hour ago
add a comment |
$begingroup$
Of course any integer can be a discrete logarithm: in a group $G$ with generator $g$, any integer $x$ is a discrete logarithm of the group element $g^x$.
Another convenient way to consider the set of discrete logarithms is as the ring $mathbf Z/nmathbf Z$, where $n$ is the order of $G$, which makes sense because $g^x = g^x bmod n$ for all $x$. This is especially convenient when $n$ is prime because then the discrete logarithms form a field.
Either way (unless the group is trivial) the discrete logarithms form a non-trivial ring with unity, which is not a group for multiplication.
answered 3 hours ago
fkraiemfkraiem
6,80021732
$endgroup$
Of course any integer can be a discrete logarithm: in a group $G$ with generator $g$, any integer $x$ is a discrete logarithm of the group element $g^x$.
Another convenient way to consider the set of discrete logarithms is as the ring $mathbf Z/nmathbf Z$, where $n$ is the order of $G$, which makes sense because $g^x = g^x bmod n$ for all $x$. This is especially convenient when $n$ is prime because then the discrete logarithms form a field.
Either way (unless the group is trivial) the discrete logarithms form a non-trivial ring with unity, which is not a group for multiplication.
answered 3 hours ago
fkraiemfkraiem
6,80021732
answered 3 hours ago
fkraiemfkraiem
6,80021732
answered 3 hours ago
fkraiemfkraiem
6,80021732
answered 3 hours ago
fkraiemfkraiem
6,80021732
6,80021732
$begingroup$
what if n is a square? If n = k^2, then k is not a discrete log mod n.
$endgroup$
– grovkin
2 hours ago
$begingroup$
@grovkin Why not? $k$ is a discrete log of $g^k$. Are you confusing it with quadratic residue?
$endgroup$
– fkraiem
2 hours ago
$begingroup$
I was just looking for an example of a group element which would not generate anything but itself. I guess copies of Z/2Z is what I should have gone with. But it's not an interesting example. What about Z/nZ, where n = p^l? What is the discrete log of 1+p^l-p^(l-1)? p is a prime, of course.
$endgroup$
– grovkin
1 hour ago
$begingroup$
@grovkin To talk about discrete logs, you need a cyclic group and a generator. If your group is Z/nZ (additive) with generator 1, the discrete log of k is k.
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– fkraiem
1 hour ago
add a comment |
$begingroup$
what if n is a square? If n = k^2, then k is not a discrete log mod n.
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– grovkin
2 hours ago
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@grovkin Why not? $k$ is a discrete log of $g^k$. Are you confusing it with quadratic residue?
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– fkraiem
2 hours ago
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I was just looking for an example of a group element which would not generate anything but itself. I guess copies of Z/2Z is what I should have gone with. But it's not an interesting example. What about Z/nZ, where n = p^l? What is the discrete log of 1+p^l-p^(l-1)? p is a prime, of course.
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– grovkin
1 hour ago
$begingroup$
@grovkin To talk about discrete logs, you need a cyclic group and a generator. If your group is Z/nZ (additive) with generator 1, the discrete log of k is k.
$endgroup$
– fkraiem
1 hour ago
$begingroup$
what if n is a square? If n = k^2, then k is not a discrete log mod n.
$endgroup$
– grovkin
2 hours ago
$begingroup$
what if n is a square? If n = k^2, then k is not a discrete log mod n.
$endgroup$
– grovkin
2 hours ago
$begingroup$
@grovkin Why not? $k$ is a discrete log of $g^k$. Are you confusing it with quadratic residue?
$endgroup$
– fkraiem
2 hours ago
$begingroup$
@grovkin Why not? $k$ is a discrete log of $g^k$. Are you confusing it with quadratic residue?
$endgroup$
– fkraiem
2 hours ago
$begingroup$
I was just looking for an example of a group element which would not generate anything but itself. I guess copies of Z/2Z is what I should have gone with. But it's not an interesting example. What about Z/nZ, where n = p^l? What is the discrete log of 1+p^l-p^(l-1)? p is a prime, of course.
$endgroup$
– grovkin
1 hour ago
$begingroup$
I was just looking for an example of a group element which would not generate anything but itself. I guess copies of Z/2Z is what I should have gone with. But it's not an interesting example. What about Z/nZ, where n = p^l? What is the discrete log of 1+p^l-p^(l-1)? p is a prime, of course.
$endgroup$
– grovkin
1 hour ago
$begingroup$
@grovkin To talk about discrete logs, you need a cyclic group and a generator. If your group is Z/nZ (additive) with generator 1, the discrete log of k is k.
$endgroup$
– fkraiem
1 hour ago
$begingroup$
@grovkin To talk about discrete logs, you need a cyclic group and a generator. If your group is Z/nZ (additive) with generator 1, the discrete log of k is k.
$endgroup$
– fkraiem
1 hour ago
add a comment |
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$begingroup$
@kelalaka Would you mind expanding upon "The discrete log is defined according to a base as the logarithm."
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– JohnGalt
4 hours ago
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@kelalaka also if "0 is not a part of the multiplicative group" does that mean that not all integers are part of a discrete log?
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– JohnGalt
4 hours ago
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Whomever down voted my question, I respect the decision, however, it would be helpful if you commented as to why you down voted it.
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– JohnGalt
3 hours ago
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Your question currently has no downvotes. That said, I'm tempted to vote to close it as unclear, since it seems to be based on some kind of a confusion of terminology, and it's literally not clear to me what you're trying to ask. The two answers already below are both correct and well written, but they answer completely different questions.
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– Ilmari Karonen
45 mins ago
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(In particular, every integer is, trivially, both a part of infinitely many different multiplicative groups and the discrete logarithm of infinitely many elements of some multiplicative group with respect to some base. And these two facts have nothing to do with each other.)
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– Ilmari Karonen
37 mins ago