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For a coprime group of N, it looks to me that every element can multiply a positive integer up to N+1 then mod N to generate the whole original coprime group. Is this true? If so, what property is this called?

For example, N = 15, and element of coprime group N will be {1,2,4,7,8,11,13,14}. I have tried all elements and it works. For example, for element 4:

4 * 2 mod 15 = 8

4 * 4 mod 15 = 1

4 * 7 mod 15 = 13

4 * 8 mod 15 = 2

4 * 11 mod 15 = 14

4 * 13 mod 15 = 7

4 * 14 mod 15 = 11

4 * 16 mod 15 = 4

So it generates the whole original coprime group N. It seems this should be some property, but I couldn't find the answer.

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  • $\begingroup$ Welcome to Math SE. If you haven't yet, you may find Wikipedia's Multiplicative group of integers modulo $n$ article interesting. Also, what is multiplicative group of all integers coprime with $N$ called? may help. I didn't suggest it as a duplicate as I'm not sure what sort of details or explanation you're looking for when you're asking about some "property". $\endgroup$ Apr 10, 2020 at 22:16
  • $\begingroup$ Thanks. Yes, the other question is more about Zn* coprime group, but my question is a little beyond that. I'm wondering for every element of Zn*, if we multiply it by an integer, it can generate the whole group again. $\endgroup$
    – 6cfc1
    Apr 11, 2020 at 16:14

1 Answer 1

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Euler employs this fact as the key lemma in his proof of the totient theorem.

Left multiplication by elements of a group always permutes the elements of the group,. It defines a group action. So you actually get a homomorphism from $G$ to $\operatorname{\\Sym}G$.

This is how Cayley's theorem, which says that every group can be embedded in a symmetric group, is done.

Since it works for any $G$, it works for the group of units, $\Bbb Z_n^×$.

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  • $\begingroup$ Thank you. The proof of Euler theorem helped. But it's a little bit different I guess. In Euler theorem it's aG congruent to G, a is fixed. But my question is more for every element G, multiply it by a positive integer, a variable, still congruent to G. $\endgroup$
    – 6cfc1
    Apr 11, 2020 at 16:17
  • $\begingroup$ In your example, you chose $4$, which is coprime to $15$. Without that it doesn't work. $\endgroup$
    – user403337
    Apr 11, 2020 at 18:07
  • $\begingroup$ Yes, the elements which satisfies this property should all from the coprime group, Zn*. My point is in my question, the element is fixed to be one from Zn*, then the multiplier is variable (2, ..., n+1). But in Euler's theorem proof, the multiplier is fixed, then multiple to every element from the group Zn*. $\endgroup$
    – 6cfc1
    Apr 12, 2020 at 16:01
  • $\begingroup$ It looked like you chose elements of $\Bbb Z_{15}^×$. At any rate, what you are asking about brought the theorem to mind. $\endgroup$
    – user403337
    Apr 12, 2020 at 16:10

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