This question is different from this question as I want to find all primitive roots, and not just some.
Is my following approach correct?
We have $121=11^2$, with $11$ an odd prime, and $2 \ge 1$, hence $\mathbb{Z}/ 11^2 \mathbb{Z}$ is cyclic.
Now we want to find a generator for $\mathbb{Z}/11^2\mathbb{Z}$. I found a suggestion for this here see the answer by Jack D'Aurizio, but it required some theory I didn't cover (Hensel's lifting lemma), so I try something different.
We have that $\phi(121) = 110 = 2*5*11$, hence we look for an element of order $110$. We pick elements from $k = 2,3,4, …$, and compute $k^n$ for $n \in \{2,5,11,2*5,2*11,5*11 \}$ if all of these give $k^n\not=1$ we know that $k$ is a primitive root modulo $121$.
Can this process of finding a generator been done more efficient? An idea would be this: suppose we find $k,l,m$ of orders $2,5,11$ respectively we know that $klm$ has order $2*5*11$ by them being coprime. Am I correct?
Now we have the following group isomorphism $\psi : (\mathbb{Z}\backslash 110\mathbb{Z},+) \rightarrow (\mathbb{Z}\backslash 121\mathbb{Z})^*$ given by: $\psi(n) = k^n$. Now a primitive root modulo 121 are precisely the elements of order $110$. As Dietrich Burde mentioned: $(\mathbb{Z}\backslash 110\mathbb{Z},+)$ is cyclic, hence it has $\phi(110) = 40$ generators.
And as isomorphisms preserve order. It suffices to find all elements of order $110$ in $(\mathbb{Z}\backslash 110\mathbb{Z},+)$. But $n \in (\mathbb{Z}\backslash 110\mathbb{Z},+)$ has order 110 IFF $\gcd(n,121) = 1$. Hence we need to find all elements in $(\mathbb{Z}\backslash 110\mathbb{Z},+)$ that are coprime with $110$.
Is there an efficient way to do this? Now given such an $n \in (\mathbb{Z}\backslash 110\mathbb{Z},+)$, we obtain the corresponding primitive root modulo $121$ via $\psi(n)=k^n$.
Best Answer
Here is one way to do it that requires knowing only a primitive root $\bmod p$ to get all primitive roots $\bmod p^2$, provided that $p$ is twice a smaller prime plus $1$, so that all nonzero residues besides $\pm 1$ are either quadratic or primitive. I will demonstrate it for $p=5$ letting you apply it for $p=11$.
Start with $2$ as a primitive root $\bmod 5$. Then there will be a series of primitive roots $\equiv 5k+2 \bmod 25$. For one residue $k \bmod 5$ we will have $(5k+2)^4\equiv 1 \bmod 25$ and that value of $k$ must be rejected, but other values of $k$ will all work.
Use the Binomial Theorem to get $(5k+2)^4$, but retain only the first ad zero powers of $k$ (why?). Thus
$(5k+2)^4 \equiv (4)(5k)(2^3)+2^4 \equiv 1 \bmod 25$
$10k+16 \equiv 1, 10k \equiv 10 \bmod 25$
Divide by $10$ noting that the modulus is to be divided by $gcd(10,25)=5$. Then $k\equiv 1 \bmod 5$ meaning $7$ will not be a primitive root $\bmod 25$ but all other residues two greater than a multiple of $5$ will work. We therefore get a subset of primitive roots:
$\{2,12,17,22\}$
Now take the root we rejected, $7$. If we multiply the above primitive roots by it, we get nonprimitive roots because these are quadratic residues. But multiply by $7$ twice, that is by $7^2$, and we get another set of primitive roots because the products are non-quadratic and also miss being $\equiv -1 \bmod 5$. Since $7^2 \equiv 24 \equiv -1 \bmod 25$ this gives another subset of primitive roots
$\{3,8,13,23\}$
Multiplying by $7^2$ again cycles us back to the first subset, so the complete set of primitive roots $\bmod 25$ will be the union of the two subsets
$\{2,3,8,12,13,17,22,23\}$.
Now apply this method to$11=2×5+1$ and see what happens. Since $11$, like $5$, is not one greater or one less than a multiple of $8$, $2$ will be a primitive root $\bmod 11$. Since there are four such primitive roots $\bmod 11$ overall you will need to generate four subsets before taking their union.