Uon request, I will clarify -- and partially take back! -- my earlier comment.

What I find completely unobjectionable is Mac Lane's definition of a subcategory $\mathcal{D}$ of a category $\mathcal{C}$ as being given by subclasses of objects and morphisms which forms a category under the induced composition. I don't see what else you would want a subcategory to be. I do agree that the notion of "subcategory" is not one of the more useful categorical concepts I know, and it even has some potential to be evil in the sense that modern categorists use the word. (Surely it would be more in the spirit of things to talk about a functor from $\mathcal{C}$ to $\mathcal{D}$ which satisfies certain "injectivity" properties.)

Now let's return to the statement "I don't want Groups to be a subcategory of Sets". In my comment I said that I did want this, but I don't now know why I said that: I think I must simply have been confused. Indeed, it is not obvious to me that this definition makes Groups a subcategory of Sets, at least not in any unique or benign way.

If you asked me to spell out the most evident categorical relationship between sets and groups, I would first of all point to the category NonEmptySets -- now that's a subcategory of Sets! -- and then the "forgetful" functor from Groups to NonEmptySets. This functor is (at least assuming the Axiom of Choice) surjective: every nonempty set is the underlying set of some group. But most sets can be endowed with a group law in multiple (usually nonisomorphic) ways, so this is not an "inclusion functor".

(Even the other way around, namely the free group functor from Sets to Groups seems not to quite make Sets into a subcategory of groups, because the class of sets is not a subclass of the class of groups.)

Maybe you are thinking of doing something tricky: defining a group to be an ordered pair [identifying ordered pairs with sets in one of the usual -- silly! -- ways] $(S,\circ)$ where $S$ is a set and $\circ$ is a subset of $S \times S \times S$ satisfying certain axioms. (Note that this is definitely incompatible with the above way of thinking about groups as having -- but not being -- an "underlying set".) But isn't this especially evil?

Comments more than welcome.

Upon request, I will clarify -- and partially take back! -- my earlier comment.

What I find completely unobjectionable is Mac Lane's definition of a subcategory $\mathcal{D}$ of a category $\mathcal{C}$ as being given by subclasses of objects and morphisms which forms a category under the induced composition. I don't see what else you would want a subcategory to be. I do agree that the notion of "subcategory" is not one of the more useful categorical concepts I know, and it even has some potential to be evil in the sense that modern categorists use the word. (Surely it would be more in the spirit of things to talk about a functor from $\mathcal{C}$ to $\mathcal{D}$ which satisfies certain "injectivity" properties.)

Now let's return to the statement "I don't want Groups to be a subcategory of Sets". In my comment I said that I did want this, but I don't now know why I said that: I think I must simply have been confused. Indeed, it is not obvious to me that this definition makes Groups a subcategory of Sets, at least not in any unique or benign way.

If you asked me to spell out the most evident categorical relationship between sets and groups, I would first of all point to the category NonEmptySets -- now that's a subcategory of Sets! -- and then the "forgetful" functor from Groups to NonEmptySets. This functor is (at least assuming the Axiom of Choice) surjective: every nonempty set is the underlying set of some group. But most sets can be endowed with a group law in multiple (usually nonisomorphic) ways, so this is not an "inclusion functor".

(Even the other way around, namely the free group functor from Sets to Groups seems not to quite make Sets into a subcategory of groups, because the class of sets is not a subclass of the class of groups.)

Maybe you are thinking of doing something tricky: defining a group to be an ordered pair [identifying ordered pairs with sets in one of the usual -- silly! -- ways] $(S,\circ)$ where $S$ is a set and $\circ$ is a subset of $S \times S \times S$ satisfying certain axioms. (Note that this is definitely incompatible with the above way of thinking about groups as having -- but not being -- an "underlying set".) But isn't this especially evil?

Comments more than welcome.