= Solution
Part (c) shows that every finitely presented <group> embeds in a finitely presented <group> with no nontrivial <finite quotients of a group>. Consequently this property also has no possible forbidden witness $A_-$. Therefore
$$
\boxed{\text{Having no nontrivial finite quotients is not a Markov property}.}
$$
Both properties have positive witnesses, for example $C_2$ and the <trivial group> respectively. It is the required embedding obstruction that fails in each case.
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