Primitive root in a finite field
You're right. In general if a $g$ is not a primitive element of $\operatorname{GF}(p)$ it generates a smaller subgroup of $\operatorname{GF}(p)^{\ast}.$ Since $|\operatorname{GF}(p)^{\ast}|=p-1$ is divisible by 2, this subgroup can be as small as 2 in size, and its size can take on any value $w,$ where $w \mid (p-1).$