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A pearl is an ordered pair $\ \mathbf P:=(G\,\ S),\ $ where $\ G\ $ is a group, and $\ S\ $ is a non-empty subset of G which does not contain the neutral element of $\ G\ $ (i.e. not 1 in the multiplicative notation, nor 0 in the additive notation).

If $\ \mathbf Q:=(H\,\ T)\ $ is another pearl then a morphism $\ f:\mathbf P\rightarrow \mathbf Q\ $ is defined as a group homomorphism $\ f:G\rightarrow H\ $ such that $\ f(S) \subseteq T$.

This defines the (general) category of pearls. Pearls for which their group is Abelian, form a full subcategory (with all their morphisms between them).

Let's define a free pearl $\ \mathbf P\ $ (as above) as one for which a subset $\ F\subseteq G\ $ is a set of the free generators of the group $\ G,\ $ and $\ S\subseteq F.\ $ Possibly, every free pearl is projective (in the category of pearls).

The definition of an abelian-free pearl (and the issue of projectivity) is similar as for the category of abelian pearls.

QUESTION   Can you provide examples of projective pearls which are not free? The same for the case of the category of abelian pearls.

REMARK This requires paying close attention to pearl epimorphisms.

                M O T I V A T I O N

1. Categories of enhanced groups are among the most important. After groups, pearls are among the simplest among them. Thus it provides a view onto the next categories which already appear;

2. There are problems on the general categories (or a general class of them) when we would like to obtain a positive result or else a counter-example. Then pearls may provide a relatively simple testing ground and they may project onto other categories.

3. Truly, the notion of pearls appeals to me, I find them attractive.

EDIT   After reading @ArturoMagidin comments, I replaced my careless obviously by careful possibly. (I'd like to see Arturo's comments posted by him as one or more answers to make them more readable, please).

A pearl is an ordered pair $\ \mathbf P:=(G\,\ S),\ $ where $\ G\ $ is a group, and $\ S\ $ is a non-empty subset of G which does not contain the neutral element of $\ G\ $ (i.e. not 1 in the multiplicative notation, nor 0 in the additive notation).

If $\ \mathbf Q:=(H\,\ T)\ $ is another pearl then a morphism $\ f:\mathbf P\rightarrow \mathbf Q\ $ is defined as a group homomorphism $\ f:G\rightarrow H\ $ such that $\ f(S) \subseteq T$.

This defines the (general) category of pearls. Pearls for which their group is Abelian, form a full subcategory (with all their morphisms between them).

Let's define a free pearl $\ \mathbf P\ $ (as above) as one for which a subset $\ F\subseteq G\ $ is a set of the free generators of the group $\ G,\ $ and $\ S\subseteq F.\ $ Obviously Possibly, every free pearl is projective (in the category of pearls).

The definition of an abelian-free pearl (and the issue of projectivity) is similar as for the category of abelian pearls.

QUESTION   Can you provide examples of projective pearls which are not free? The same for the case of the category of abelian pearls.

REMARK This requires paying close attention to pearl epimorphisms.

                M O T I V A T I O N

1. Categories of enhanced groups are among the most important. After groups, pearls are among the simplest among them. Thus it provides a view onto the next categories which already appear;

2. There are problems on the general categories (or a general class of them) when we would like to obtain a positive result or else a counter-example. Then pearls may provide a relatively simple testing ground and they may project onto other categories.

3. Truly, the notion of pearls appeals to me, I find them attractive.

EDIT   After reading @ArturoMagidin comments, I replaced my careless obviously by careful possibly. (I'd like to see Arturo's comments posted by him as one or more answers to make them more readable, please).