2
$\begingroup$

I know that seems a dumb question, but I want to make sure.

  • 2
    Sometimes people write $\mathbb{Z}_p$ to mean the $p$-adic integers (http://en.wikipedia.org/wiki/P-adic_integer ). These don't form a finite group.2012-09-19

2 Answers 2

2

Yes, $\mathbb{Z}_p$ is the cyclic group of order $p$; that is, it is generated by a single element of order $p$. We get $\mathbb{Z}_p$ by forming a group from the equivalence classes of the integers modulo $p$.

  • 0
    @MJD Yes, my mistake. Good catch.2012-09-19
1

I am writing here maybe it completes Tarnation's answer.

Definition: Let $n$ is a fixed positive integer number and $h, k$ are two arbitrary integers so the number $r$: $h+k=nq+r, 0\leq r\leq n$ is called the sum of $h$ and $h$ modulo $n$. Now consider the set: $\{1,2,3,...,n-1\}$ then:

Theorem: the above set forms a cyclic group, called $\mathbb Z_n$, under addition modulo $n$.

Of course if $h+k=r$ happens in group $\mathbb Z_n$ then for common addition in $\mathbb Z$ we see $h+k\equiv r$ (mod $n$). The order of set $\mathbb Z_n$ is finite so it is a finite group of order $n$. In you problem if $n=p$ a prime number, as Tarnatian's second comment, all elements of $\mathbb Z_p$ could generate the whole group.