The elements of $\Bbb{Z}_{20}^{\times}$

Question

The elements of $\Bbb{Z}_{20}^{\times}$, as I understand, are all the number from 1 to 20 included that are relatively prime to 20? I am having troubles finding a coherent definition of this kind of groups... I thought it should be: $\Bbb{Z}_{20}^{\times}=${$1,3,7,9,11,13,17,19$} but it is written somewhere that $\Bbb{Z}_{20}^{\times}=${$\pm 1,\pm 3,\pm 7,\pm 9$}, which is somehow rational because the cardinality is the same, also because for every $g$, $-g$ might stand for the inverse of $g$ , it is in that case $20-g$, which would explain replacing $11$ with $-9$, but how come? This is a multiplication group and the addition is not relevant. I am confused and would appreciate you direct explanation...

Answer 1

Though it is not false, the notation $\pm a$ is rather used, when convenient, for $\mathbf Z_p$, $p$ a prime number.

In the specific case, you must understand that $ \mathbf Z /20 \mathbf Z$ is aquotient ring of $\mathbf Z$, and as such, there is defined an addition and a multiplication of its elements. $(\mathbf Z /20 \mathbf Z)^\times$ denotes the units of this ring, i.e. the elements which have a multiplicative inverse. An element such as the residue class of $11$ actually is the set of integers $11+20\mathbf Z$, which differ from $11$ by a multiple of $20$, so $11+20\mathbf Z$ is indeed the same set as $-9+20\mathbf Z$.

Answer 2

Many times it is a great advantage working with negatives. Suppose a train starts at 9 pm and takes 23 hours to reach its destination. To find the time of arrival what does one do? add 9+23 and take reminder wrt 24? We take short cut and caluclate 9-1= 8pm Because 23 is -1 mod 24. To prove the famous quadratic reciprocity law, Gauss used negatives while working modulo a prime.

Answer 3

I believe that the problem lies in the fact that the subtle distinction between

1. the set of cosets $\{x+20\mathbb{Z}:x\in\mathbb{Z}\}$ with multiplication defined by $$(x + 20\mathbb{Z})\circ(y + 20\mathbb{Z}) = x y + 20\mathbb{Z}$$

2. the set $\{1,3,7,9,11,13,17,19\}$ and multiplication defined by $$x \circ y =(x \cdot y) \% 20$$ where $\cdot$ is the multiplication of $x, y$ "as integers" and $\%$ is the modulus operator

is so often mixed up (well, of course, they are essentially the same).

If you use the first definition, for example $17 + 20 \mathbb{Z}$ simply is the same set of integers (coset) as $-3 + 20 \mathbb{Z}$. Because two cosets $x + 20\mathbb{Z}$, $y+20\mathbb{Z}$ are equal if for its representatives $x, y$ the following holds: $$ x \equiv y \pmod{20}\, .$$ You can write $\overline{-3}$ or $\overline{17}$ for the corresponding cosets of $20\mathbb{Z}$, then $\overline{-3} = \overline{17}$ and $$\{\overline{1},\overline{3},\overline{7},\overline{9},\overline{11},\overline{13},\overline{17},\overline{19}\} = \{\overline{\pm 1}, \overline{\pm 3}, \overline{\pm 5}, \overline{\pm 9} \}$$ Not writing the overlines is a bit sloppy, because the elements aren't integers, they are cosets.

The second definition is inflexible. Strictly speaking $-3$ is not an element of this group. The right way (which borders on the pedantic) would be to define the multiplication for the set $\{\pm 1, \pm 3, \pm 5, \pm 9\}$ again and then show that this group is isomorphic to the group of the second definition.