How to check if a number is a generator of a cyclic multiplicative group

Suppose I have a 2048 bit prime number p. Now for the group $Z_p$, could someone please tell me an efficient algorithm to check whether a randomly chosen number is a generator for the group or not

Preliminary: $\mathbb Z_p$ is not a group under multiplication modulo $p$, for it contains the neutral element for addition modulo $p$, and that has no multiplicative inverse modulo $p$. What's wanted is an algorithm to check if $g\pmod p$ is a generator of the multiplicative subgroup of $\mathbb Z_p$, noted $\mathbb Z_p^*$, which has $p-1$ elements.

To determine that, we want to find the full factorization of $p-1=\prod{p_i}^{r_i}$. Then the desired necessary and sufficient condition is that $g\bmod p\ne0$, and for each prime $p_i$ dividing $p-1$ it holds $g^{(p-1)/p_i}\bmod p\ne1$. That easily translates to an algorithm.

Note: when $p=2$ there is no prime $p_i$ dividing $p-1$.

Note: When $p$ is a safe prime, the condition can be simplified to $g\bmod p\,\not\in\{0,1,p-1\}$ and $g^{(p-1)/2}\bmod p\,=p-1$. The corresponding algorithm is simpler.

In addition to fgrieu's correct answer, even if you don't know the factorization of $p-1$, you can still get an answer that has a high probability of being correct (assuming that the value you're testing $g$ was chosen randomly and not maliciously).

Here's how you do it: you obtain a partial factorization of $p-1$, finding all small prime factors $q_0 = 2, q_1, q_2, ...$ which are less than $\lambda$, that is $p-1 = q_0 q_1 q_2 ... q_n C$ where $C$ has no factors $< \lambda$ (and is presumably composite - if it is prime or one, you have a complete factorization and can use fgrieu answer, which always gives you the right answer). The Elliptic Curve Method of factorization can give you such a partial factorization with $\lambda \approx 2^{128}$ being practical, and even larger $\lambda$s are possible (depending on your computing resources).

Once you have that, you compute (just as fgrieu stated) $g^{(p-1)/q_i} \bmod p$, and if they are all something other than 1, then you have a generator, with a maximum probability of being wrong (assuming $g$ was chosen randomly) $< \frac{\log p} {\lambda \log \lambda}$ - for $\lambda$ in the range that ECM allows you, this may be adequately small...