= 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 becomes
$$
\bar b\bar a\bar b^{-1}=\bar a^2.
$$
Iterating <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 hence
$$
n\mid 2^n-1.
$$
The 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>.
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