= Solution
\b[False.] Take the finite, hence countable, <cyclic group> $C_6=\langle a\mid a^6=1\rangle$. Its elements $a^3$ and $a^2$ have orders two and three. A <group embedding> preserves the order of every element, whereas <conjugate group elements> have the same order: $(s^{-1}xs)^r=s^{-1}x^rs$. Therefore their images cannot be conjugate in any containing <group>. This gives the required obstruction. The <torsion-free group> hypothesis in Question 5 avoids this conflict of element orders.
Back to article page
