How to find generators in $U(n)$?

Question

For e.g- in $U(10)=\{1,3,7,9\}$ are elements and $3$ & $7$ are generators but for a big group like $U(50)$ do we have to check each and every element to be generator or is there any other method to find the generators?

Answer 1

Your question is kind of ambiguous as it's been worded currently, but I assume that you're asking about the generators of $\mathrm{U}(n)$, i.e. the group of elements of $\mathbb{Z_n}$ that have a multiplicative inverse.

Sometimes a generator doesn't exist at all. In fact, when a generator exists, it's called a primitive root and there's a classification of all $n$'s for which $\mathrm{U(n)}$ is cyclic, i.e. a generator exists:

$$n=2,4,p^k, 2p^k$$

where $p$ is an odd prime number. The number of generators, if any exists, is then $\varphi(\mathrm{U}(n))=\varphi(\varphi(n))$.

If you're looking for a constructive way to find a generator, please look here for the case where $n=p^k$ and you already know a generator for $p^{k-1}$. In fact, if you read the article, it also talks about other cases like $n=2p^k$ as well. Therefore, it answers your question to a great deal. And also check the proofs for the classification theorem I mentioned. Some of them are constructive and can give you ideas about how to construct a primitive root. Also, take a look at this Wikipedia entry.

One more thing, if you find a generator of $\mathrm{U}(n)$, let's call it $g$, then every element in the group can be written as $g^i$ for some $i\in \{0,1,\cdots,\varphi(n)-1\}$. You can find other generators of the group by noting that $h=g^j$ will be a generator when $(j,\varphi(n))=1$ and therefore it suffices to find only one generator of the group because the rest can be found from there.