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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.

JohnGalt
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  • @kelalaka Would you mind expanding upon "The discrete log is defined according to a base as the logarithm." – JohnGalt Apr 17 '19 at 18:28
  • @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? – JohnGalt Apr 17 '19 at 18:29
  • Whomever down voted my question, I respect the decision, however, it would be helpful if you commented as to why you down voted it. – JohnGalt Apr 17 '19 at 19:33
  • 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. – Ilmari Karonen Apr 17 '19 at 22:11
  • (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.) – Ilmari Karonen Apr 17 '19 at 22:19
  • @IlmariKaronen My comment was about an earlier down vote. With regard to "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" Is that statement true for 0 which is an integer? This gets to the heart of what I agree was an ill-formed question. Which after our discussion, I'd be happy to delete if you think it would be in the best interest of the community. A better question would've been can 0 be a discrete log? Which I believe the answer is no. – JohnGalt Apr 17 '19 at 22:34
  • @IlmariKaronen and to further clarify: I was trying to differentiate an integer which is "discrete" versus not all integers can constitute a discrete log. – JohnGalt Apr 17 '19 at 22:42
  • In any cyclic group $G$ with generator $g$, $0$ is a discrete log of $g^0$, which is the identity element of $G$. – fkraiem Apr 17 '19 at 23:35
  • Why does it matter? What are you worried about communicating imprecisely near the exact boundaries of where the term ‘discrete logarithm’ might or might not be applicable? In principle the only property you need is power-associativity of a binary operation on a set with a discrete topology for it to be meaningful, but that does so little to narrow down the space of possible scenarios that it's hard for me to imagine why this question even arises. We even use it in contexts like elliptic curve scalar multiplication which aren't even written as exponentiation per se. – Squeamish Ossifrage Apr 17 '19 at 23:52
  • Pick any multiplicative group, and let $b$ be an element of it. What's $b^0$? What is the (smallest non-negative) base $b$ discrete logarithm of $b^0$ in that group? (Also, I can't make any sense of your "further clarification": it seems to be missing a word somewhere, and the idea of "an integer which is discrete" seems meaningless, since discreteness is a property that describes a collection of things. You might as well ask for a syllable that rhymes, a note that is harmonic, an atom that is liquid or a person who is densely populated.) – Ilmari Karonen Apr 17 '19 at 23:55
  • @fkraiem My question may be nonsensical and/or 2) lack context in that I was trying to grok the setup of DH, which as you know deals with discrete logs, the discrete log problem, multiplicative groups, cyclic groups, etc. In someone not trained (that'd be me in cryptography) I was confused by the use of "discrete" in those terms and especially resolving the term discrete as it applies to integers versus the term used in DL and DLP. I ran across statements that all integers are discrete logs which was confusing and my question was trying to clarify. I failed :) – JohnGalt Apr 18 '19 at 01:14
  • @SqueamishOssifrage Why does it matter? Because I was trying to resolve how a statement like "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." is true....fkraiem showed an example of where it is true and in the end as you stated who cares because there's no use case for it. – JohnGalt Apr 18 '19 at 01:23
  • @IlmariKaronen RE: "further clarification" It was an attempt to answer some further questions which came up from this question. I appreciate your patience and expertise and although my question was convoluted, I've learned a bunch from the exchange and hope others can get some benefit as well. – JohnGalt Apr 18 '19 at 01:27
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    @JohnGalt In the statement you quoted, the context is discrete logs in $\mathbb Z/n\mathbb Z$ for some $n$. It means: For any integer $x$, there exists some $n$ and some $g, h \in \mathbb Z/n\mathbb Z$ such that $x = \log_g h$, i.e. $h = g^x$. To interpret that sentence, there is no need to extend the term ‘discrete log’ to the ring of all integers, or to extend the term to apply outside the context of any particular group, because the term is only ever being used within some particular group $\mathbb Z/n\mathbb Z$. – Squeamish Ossifrage Apr 18 '19 at 01:47
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    By the way, the statement you quoted is indeed not true for $0$, in the context of groups of the form $(\mathbf Z/n\mathbf Z)^*$. $0$ is indeed not part of any such group (or it is part of exactly one such group, if you allow the degenerate case $n=1$), and it is a discrete log of exactly one element of any such group (the identity element), not infinitely many. It is however a discrete log of an element, so it is a discrete log indeed. – fkraiem Apr 18 '19 at 02:59
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    All integers between $1$ and $k−1$ will be results of some $log_b a$, where $b$, $a$ are in $Z/nZ$ and (this is the important part) $k$ is the order of the highest-order element of the multiplicative group, $(Z/nZ)^{\times}$, of units of $Z/nZ$. So, for example (mod 9), 5 is a discrete log bc $4^5=7$(mod 9), but 7 is not a discrete log (mod 9) because $(Z/9Z)^{\times}$ has only 6 elements. So no element of $(Z/9Z)^{\times}$ can have order 7. – grovkin Apr 18 '19 at 06:09

2 Answers2

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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 $a \not\in G'$ 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 $F\backslash\{0\}$ they are forming a cyclic group under multiplication. For a proof see the Theorem 1.

kelalaka
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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 $\mathbf Z/n\mathbf Z$, 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.

fkraiem
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