Can you give me an example of generator of multiplicative group $$(\mathbb{Z}/p\mathbb{Z})^{*}=\{1, 2, \ldots, p-1\}.$$
Thanks.
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41 for $p=2$ is an example. – Phira Nov 06 '11 at 00:18
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Depends on what $p$ is. – Zev Chonoles Nov 06 '11 at 00:18
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@Zev Chonoles: is there exists clear formula for generator for any prime $p$? – Aspirin Nov 06 '11 at 00:22
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1Finding a generator for this group is computationally difficult in general. – Bill Cook Nov 06 '11 at 00:23
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@Aspirin: No. – Zev Chonoles Nov 06 '11 at 00:23
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No. There is no clear formula. – Bill Cook Nov 06 '11 at 00:24
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@Bill Cook: and for $p=4k+1$? P.S. I want to proof that in such group exists element of order $4$. – Aspirin Nov 06 '11 at 00:28
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1@Aspirin: Why not make that your question... – Zev Chonoles Nov 06 '11 at 00:29
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@Aspirin, IIRC the multiplicative group is always cyclic. For an existence proof that should be enough. – hmakholm left over Monica Nov 06 '11 at 00:32
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@Zev Chonoles: because if generator there exists, than least statement will be obviously – Aspirin Nov 06 '11 at 00:33
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@Henning Makholm: I'm very sorry but why this group is cyclic? – Aspirin Nov 06 '11 at 00:36
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You are looking for primitive roots mod $p$. See Sloane's sequence A001918, http://oeis.org/A001918. – Hans Engler Nov 06 '11 at 00:37
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@Aspirin, it's a general fact about all finite fields that their multiplicative groups are cyclic. I don't remember a proof or reference, though, sorry. – hmakholm left over Monica Nov 06 '11 at 00:38
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@Henning Makholm: Am I right that multiplicative group of ahy field always cyclic? – Aspirin Nov 06 '11 at 00:39
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@Aspirin: See this question for a proof that $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic. – bzc Nov 06 '11 at 00:41
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@Aspirin, last I checked it wasn't the case for $\mathbb Q$ or $\mathbb R$. – hmakholm left over Monica Nov 06 '11 at 00:41
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@Henning Makholm: Thanks a lot, a read it proof in book – Aspirin Nov 06 '11 at 00:42
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1http://en.wikipedia.org/wiki/Primitive_root_modulo_n – anon Nov 06 '11 at 00:42
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A slight generalization of that argument shows that any finite subgroup of the multiplicative group of a field is cyclic. – bzc Nov 06 '11 at 00:42
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@Henning Makholm: *for finite field – Aspirin Nov 06 '11 at 00:42
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1You should make the title of your post a question or similar statement; as it is, your title is very uninformative. – Dan Drake Nov 06 '11 at 02:29
1 Answers
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I am putting together the question with the comments to figure out the real question. And, this is the answer.
As Henning said, the group is always cyclic. This is a special case of the fact that the multiplicative group of units from a finite field is always cyclic. In a cyclic group of order $n$, there is always an element of order $d$ for any $d$ that divides $n$.
The group $\mathbb{Z} / p \mathbb{Z}$ is order $p-1$. So, if $p = 4k+1$, as you mentioned in a comment, then this group is order $4k$. In that case, 4 divides the order of the group so there is guaranteed to be an element of order 4.
GeoffDS
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I wonder if finding an efficient algorithm for the generator would imply an efficient algorithm for the discrete logarithm problem? This is what comes to mind right away. – pki Nov 06 '11 at 01:30
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2@pki, why would it? The cryptographic applications of discrete logarithms are usually for moduli where a generator is known and public already. – hmakholm left over Monica Nov 06 '11 at 01:46