 Solution

Write $\bar{a}=f(a)$, $\bar{t}=f(t)$, and $\bar{b}={\bar{a}}^{\bar{t}}={\bar{t}}^{-1}\bar{a}\bar{t}$. The elements $\bar{a}$ and $\bar{b}$ are conjugate group elements, so they have the same order of a group element, say $n$. The defining relation becomesIterating conjugation gives ${\bar{b}}^j\bar{a}{\bar{b}}^{-j}={\bar{a}}^{2^j}$. Since ${\bar{b}}^n=1$, taking $j=n$ yields $\bar{a}={\bar{a}}^{2^n}$, and henceThe stated consequence of Fermat's little theorem implies $n=1$. Thus $\bar{a}=1$, and the surjective group homomorphism $f$ shows that $Q$ is generated by $\bar{t}$. Therefore $Q$ is a cyclic group.