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In the book Programming bitcoin by Jimmy Song, the author provides an exercise for learning about finite fields:

for k = 1, 3, 7, 13, 18, what is this set in F(sub)19?
{k * 0, k * 1, k * 2, k * 3, .... k * 18}
Do you notice anything about these sets?

and then comments:

The answer to Exercise 5 is why we choose to use fields
with a prime number of elements. 
No matter what k you chose, as long as it's greater than 0, 
multiplying the entire set by k will result in the same set as you started with

If I multiply with e.g. 3 then I get {0, 3, 6, etc}, a different set

What does the author mean by "same set as you started with"?

Henry
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    I think there might be something missing from your setup of the problem. Perhaps the "multiplication" involved here is that of arithmetic modulo $19$? Then you would have equal sets. – hardmath Jul 27 '19 at 02:08
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    Keep going with your $0,3,6..$ example making sure to reduce mod 19. They don't need to be in same order to be same set. – coffeemath Jul 27 '19 at 02:08

1 Answers1

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It isn't a different set: you get the same elements back, but in a different order. It looks like $1$ and $2$ were skipped, for instance, but $3\cdot 13=39\cong1\pmod{19}$ and $3\cdot 7=21\cong2\pmod{19}$.

Evidently, multiplication by a nonzero $k$ gives a bijection from $\Bbb F_p\to\Bbb F_p$. (For $k$ is invertible in $\Bbb F_p$.)