= Solution
A <primitive root> modulo $N$ is a unit of order $\varphi(N)$; a nonzero <quadratic residue> modulo an odd prime $p$ is a square.
If a primitive root $g$ modulo $p$ were $x^2$, then its order would divide $(p-1)/2$, the order of the <subgroup> of squares, contradicting $\operatorname{ord}_p(g)=p-1$.
Modulo $11$, $2$ has order $10$, so the primitive roots are $2^k$ for $\gcd(k,10)=1$:
$$
2,\quad 8,\quad 7,\quad 6.
$$
Finally $2^{10}=1024\equiv56\not\equiv1\pmod{121}$, so the lifting criterion makes $2$ primitive modulo $121$. It is plainly the smallest possible positive primitive root.
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