How to systematically find all the elements (ℤ/19ℤ)∗ with order 6 in the form of $2^a$ where $0≤a≤18$? So my first thought about this question is to find element such that d=gcd(a,18)=3. And this equation gives a = 3, 12, and 15. Hence $2^{3}$, $2^{12}$ and $2^{15}$ are elements of (ℤ/19ℤ)∗ with order 6. Is it correct?
Also, with the same reasoning, we can get $2,2^{5},2^{7},2^{11},2^{13}$ and $2^{17}$ are generators of (ℤ/19ℤ)∗ in the form of $2^a$ where $0≤a≤18$. Thank you!
Since $2^9\equiv 18\equiv-1\pmod{19}$, we know that $2$ is a generator of the group $(\mathbb{Z}/19\mathbb{Z})^*$, which is cyclic.
Therefore $2^3$ is a generator of the unique six element subgroup. If $g$ is a generator of a six element cyclic group, then the other generator is $g^5$. Thus the elements you're looking for are $2^3$ and $2^{15}$.
Note that $2^{12}$ hasn't order $6$, because it has order $3$.
You are correct that the generators of $(\mathbb{Z}/19\mathbb{Z})^*$ are $2$, $2^5$, $2^7$, $2^{11}$, $2^{13}$ and $2^{17}$.
