How can I find the generator of a composite group and $Z_p*$? I was doing some research on elliptic curves. I know how to find the generator of $Z_p$ (this is a prime group). But I came across the term $Z_p*$ (group containing elements that relatively prime to $p$, which is composite obviously). 

So I want to know how to find the generator of a composite group. How can I find the "generator" of a "composite group" (group order is composite)? And how can I find the "generator of $Z_p*$" (field with composite elements ranging from $\{1,2,\dots,p-1\}$)?
$\newcommand{\Z}{\mathbb{Z}}$If I have understood your question correctly, your goal is to find a primitive root modulo $p$, also called a generator of $(\Z/p\Z)^\times$, knowing that $p$ is prime.

Do you know the prime factorization of $\phi(p) = p - 1$?  If you don't, this is hard.  If you do, there's at least one common fast case, and there's always a general slow case.


Fast case.  Is $p$ a safe prime—that is, is there another prime $q$ such that $p = 2 q + 1$?  If so, then there are only four possible orders, $\{1,2,q,2q\}$, corresponding, respectively, to the subgroups $\{1\}$, $\{1, -1\}$, the quadratic residues, and the whole group.  Thus to find a generator of $(\Z/p\Z)^\times$ it suffices to find a quadratic nonresidue other than $-1$.

You can use the law of quadratic reciprocity to quickly pick a generator.  For example, if $p \equiv 3 \pmod 8$ or $p = 5$, then $2$ is a quadratic nonresidue and hence a generator; otherwise $p \equiv 7 \pmod 8$, since $p \equiv 3 \pmod 4$ by virtue of being a safe prime above 5, so $-2$ is a quadratic nonresidue and hence a generator.

(Why is this case common?  Often the goal is to find a generator for a Diffie–Hellman group, which over finite fields is always done with a safe prime modulus—although in that case usually one seeks a generator of the order-$q$ subgroup instead, i.e. a quadratic residue other than $-1$.  See, e.g., RFC 2412, Appendix E ‘The Well-Known Groups’.)
Otherwise, general case.  Let $\phi(p) = p - 1 = q_0^{e_0} q_1^{e_1} \cdots q_{k-1}^{e_{k-1}}$ for primes $q_i$.  March through the quadratic nonresidues $x \in (\Z/p\Z)^\times \setminus \{-1\}$, and for each distinct factor $q_i$ of $p - 1$, check whether $$x^{\phi(p)/q_i} \equiv 1 \pmod{p}.$$  If for some $x$ the powers $x^{\phi(p)/q_i}$ are all not congruent to 1 modulo $p$, then you have found an element of maximal order $\phi(p)$ which is therefore a generator.

(It doesn't hurt to test all elements, but you may be able to skip quadratic residues faster than computing the modular exponentiation.)