= Solution
{c}
The extra <Pauli gate> on the control is removed by $Z_aG=C_Z$. Conjugating the target by a <Hadamard gate> then gives
$$
\boxed{C_X=(I\otimes H)(Z\otimes I)G(I\otimes H).}
$$
Thus, from left to right in the network, apply $H_b$, then $G$, then $Z_a$ and $H_b$; the last two gates commute.
\Image[/past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2021-ion-gate-networks.png]
{title=Synthesis of a <CNOT gate> from the ion gate $G$, and preparation of a <Bell pair>}
For a direct proof, the control-zero block of $G$ is $I_b$ and its control-one block is $-Z_b$. Multiplication by $Z_a$ changes the latter into $Z_b$. Since $H_bI_bH_b=I_b$ and $H_bZ_bH_b=X_b$, the result is
$$
|g\rangle\langle g|_a\otimes I_b+|e\rangle\langle e|_a\otimes X_b,
$$
which is exactly the <CNOT gate> on all inputs, including <quantum superpositions>.
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