Is $\mathbf{Grp}$ a concrete category?

Question

Is $\mathbf{Grp}$ a concrete category? I thought it is, but then the group of symmetries of a square and the quaternion group are both of the order 8, and they are not isomorphic as groups. But sets of the same cardinality are isomorphic as sets, so a standard forgetful functor $(G, \cdot) \mapsto G$ is not faithful! Am I missing something here?

Answer 1

Yes, you have the wrong definition of a faithful functor. A functor F is faithful precisely when F sends morphisms injectively, but nowhere is there required any kind of injectivity on objects.