I know that seems a dumb question, but I want to make sure.
Is it safe to say that $Z_p$ is always a finite group? And that the order of $Z_p$ is p?
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2It might be a good exercise for you to verify the group axioms on the set $\{0,1,\dots,p−1\}$ with the operation of addition modulo $p$ for prime $p$. – 2012-09-19
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2Sometimes 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
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 $p$.
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1But if $p$ is prime, it is generated by every single element right(except the identity)? – 2012-09-19
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1That is correct. If the order is $n$, the group is generated by elements of order relatively prime to $n$. – 2012-09-19
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2@Tarnation You misspoke. The group is *generated* by elements of **order** $n$; these are exactly the elements of $\{0, 1,\ldots, n-1\}$ which are relatively prime to $n$. – 2012-09-19
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0@MJD Yes, my mistake. Good catch. – 2012-09-19
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.