It is well known that the existence of an inaccessible cardinal implies the consistency of ZFC. However, I am curious about the converse. Does the consistency of ZFC (plus ZFC) imply the existence of an inaccessible cardinal? If we suppose ZFC is consistent, then it has a model. Could that model be used to construct an inaccessible cardinal? Could perhaps the cardinality of that model itself be an inaccessible cardinal?
The answer is no.
Suppose there is an inaccessible cardinal. We will construct a model of ZFC + Con(ZFC) + "There is no inaccessible cardinal".
Let $\kappa$ be the smallest inaccessible cardinal. It is routine to verify that $V_\kappa\models ZFC + $ "There is no inaccessible cardinal".
Let $X\preceq V_\kappa$ be countable. Take the Mostowski collapse $M\cong X$. As $M$ is countable and transitive, $M\in V_\kappa$. By elementarity, $X\models ZFC$, so $M\models ZFC$, so $V_\kappa\models (M\models ZFC)$, so $V_\kappa \models $ Con (ZFC).