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Is it true that if p is an odd prime number, then there exists some integer a such that the permutation of Ф(p) induced by multiplication by a mod p consists of exactly 2 cycles?

Jingting931015
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  • What do you mean by "...the permutation of $\Phi(p)$ induced by multiplication by $a\pmod{p}$"? – Servaes Nov 01 '18 at 16:15
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    It means that the function of the permutation of the set Φ(p) is multiplication of a (mod p). Φ(p) is the set of numbers between 0 and p-1 which are coprime with p (In this case, it's just the set of numbers between 1 and p-1 since p is prime) – Jingting931015 Nov 01 '18 at 16:23
  • Ah, the set you denote $\Phi(p)$ is commonly denoted $(\Bbb{Z}/p\Bbb{Z})^{\times}$ or $(\Bbb{Z}/p\Bbb{Z})^*$. This notation comes from ring theory. Could you please include this clarification in the question itself? Also, are $n$ and $a$ supposed to be the same integer? – Servaes Nov 01 '18 at 16:28
  • And interesting question! What are your thoughts on the problem? What have you tried so far? – Servaes Nov 01 '18 at 16:31

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It is a basic fact of group theory that for every odd prime number $p$ the group $\Phi(p)=(\Bbb{Z}/p\Bbb{Z})^{\times}$ is cyclic of order $p-1$. For a proof, see this question. So if $g$ is a generator, then multipication by $g$ is represented by a $p-1$-cycle, and hence multiplication by $g^2$ is represented by its square, the product of two disjoint $\frac{p-1}{2}$-cycles.

Servaes
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  • Sorry that I have not studied group theory yet. I wonder if I understand this correctly: by generator do you mean a single number the mulplication of which can produce the entire group? So if we choose the number a to be the square of g, which is the generator, we get a permutation consisting of two cycles. – Jingting931015 Nov 02 '18 at 04:06
  • Yes, that is what I mean by generator. Note that in general, there are many different generators; if $g$ is a generator and $k$ is coprime to $p-1$, then $g^k$ is also a generator. And indeed, the square of any generator yields a permutation consisting of two cycles. – Servaes Nov 02 '18 at 14:28