I am trying to find all incongruent primitive roots of $14$. I know that there are $\phi{(\phi{(14)})}=2$ incongruent primitives roots, where I have found that both $3$ and $5$ are primitive roots mod $14$.
My question is: why aren't $11$ and $13$ also primitive roots mod 14? Both $11^6 \equiv 13^6 \equiv 1$ mod $14$, by Euler's Theorem. So why are there only $2$ incongruent solutions mod $14$.