Show that the only elements equal to their own inverses are $1$ and $p-1$.

Question

Let $p$ be a prime number and $U_p$ be the Abelian group of numbers $\{1, 2, 3, 4, \dots, p-2, p-1\}$ where the binary operation is multiplication $\mod p$. Show that the only elements that are equal to their own inverse are $1$ and $p-1$.

So because its abelian it is equal to its identity element so $p$ will divide $p-1$ and $1$? A nudge in the right direction would be appreciated! Thanks.

Answer 1

$q\in U_p$ is its own inverse if $q^2 = 1$, that is, if $q^2 \equiv 1$ (mod $p$). This means that $p$ divides $q^2-1 = (q+1)(q-1)$, and therefore either $p\big|(q-1)$ or $p\big|(q+1)$.

If $q\in U_p\setminus\{1,p-1\}$, then $q+1\in\{3\ldots p-1\}$ and $q-1\in\{1\ldots p-3\}$. $p$ is prime, and therefore it can only divide its multiples, and none of $\{1,2,\ldots,p-1\}$ are. This proves that no such $q$ exists.

Answer 2

Hint: If $a^2 \equiv 1$ (mod $p$), then $a^2 - 1 = (a + 1)(a-1) \equiv 0$ (mod $p$).