Prove that an element $\bar{x}$ is its own inverse iff $\bar{x} = 1,-1$

Question

Show that in $\mathbb{Z_p^\star}$ (where p is prime under the binary operiation multiplication), an element $\bar{x}$ is its own inverse if and only if $\bar{x} = \bar{1},\overline{p-1}$

This question was in my last homework, and my solution was as follows:

$\overline{x^2} = \bar{1}$ $\iff \overline{x^2-1} = \bar{0} \iff \overline{(x-1)(x+1)} = \bar{0}$

I then concluded that either $\overline{x-1}$ or $\overline{x+1}$ is $\overline{0}$ however, the marker stated I needed more justification from the of:

$\iff \overline{(x-1)(x+1)} = \bar{0}$

to

either $\overline{x-1}$ or $\overline{x+1}$ is $\overline{0}$

I asked my lecturer at the end of one lecture and he said that you need to justify that either $p$ divides $\overline{x-1}$ or p divides $\overline{x+1}$, but I don't understand how p divides either, or how that justifies anything. Could someone please explain?

@labbhattacharjee I don't get how from this step you see that $p$ is dividing anything — Warz, Feb 08 '14 at 16:47

Bobby · Accepted Answer · 2014-02-08T16:53:28.887

6

Assume that $ \overline{(x-1)(x+1)} = \bar{0}$ and by definition of $\mathbb Z_p^*$, we have $p\,|\,(x-1)(x+1)$. Thus, either $p\,|x+1$ or $p\,|\,x-1$ and it implies that $\bar{x} = \bar{1},\overline{p-1}$.

The converse is clear.

Note that $\overline 1=\{\ldots,-p+1,p+1,2p+1\ldots\}$ or $\overline 0=\{,\ldots,-p,0,p,2p,\ldots\}$. When we write $ \overline{(x-1)(x+1)} = \bar{0}$, we mean $(x-1)(x+1)=kp$ where $k\in \mathbb Z$.

edited Feb 08 '14 at 16:53

answered Feb 08 '14 at 16:46

Bobby

7,596

How did you get $p | (x-1)(x+1)$? – Warz Feb 08 '14 at 16:50
@Warz This is the definition of modulo: $a=b$ iff $p|(a-b)$ – LeeNeverGup Feb 08 '14 at 16:53
NOw, I think that it is clear for you. – Bobby Feb 08 '14 at 16:54
"Thus, either ..." Isn't this the whole point of the question? Most likely, what is being asked is to justify precisely this step: That if a prime $p$ divides a product, then it divides one of the factors. Nowhere in what is written one sees the reason why $p$ being prime is relevant. – Andrés E. Caicedo Feb 08 '14 at 16:56

score 1 · Answer 2 · answered Feb 08 '14 at 16:50

You're considering integers modulo $p$ under multiplication - so they're refering to an integer being equivalent to 0 mod $p$ if it is a multiple of $p$.

Then since $(x+1)(x-1) = 0$ in $\mathbb{Z}_p^*$, have that (in $\mathbb{Z}$) $p$ divides $(x+1)(x-1)$. Then either $p$ divides $(x+1)$ or $p$ divides $(x-1)$ - since it's a standard property that for $p$ prime, $p \vert ab \Rightarrow p \vert a \ or \ p \vert b$.

Alternatively you could quote that $\mathbb{Z}_p$ in an integral domain, but if they were expecting you to do that I think what they'd said about justifying it would've been different (so you may not have done that sort of thing yet?).

no we haven't yet, but with your reply and bagen's I understand, thanks. — Warz, Feb 08 '14 at 16:55

Prove that an element $\bar{x}$ is its own inverse iff $\bar{x} = 1,-1$

2 Answers2