Clarification need for the meaning of "underlying set" in the context of concrete categories in Hungerford's graduate Algebra text. For the following example taken from Hungerford graduate Algebra text, can someone provide clarification where the example says:
"However, in the third example after Definition 7.1, is the function $\sigma$ assigns to the group $G$ that usual underlying set $G$, then the category in question is not a concrete category(since the morphisms are not functions on the set $G$)."  In the context of the third example about multiplicative group $G$, I don't know what is considered of the underlying set, specifically is there a function $\sigma$ for the case of the multiplicative group, and as a result how is not considered as a concrete category.  I feel like Hungerford is short on details in this particular example.

Example:appears after Definition 7.6, pg 55 $G$, then the category in question is not a concrete category Definition 7.6 pg 55  The category of groups, equipped with the function that assigns to each group its underlying set in the usual sense, is a concrete category.  Similarly the categories of abelian groups and partially ordered sets, with the obvious underlying sets, are concrete categories.  However, in the third example third example, multiplicative $G$, pg 53 after Definition 7.1, definition 7.1, pp 52-53.  is the function $\sigma$ assigns to the group $G$ that usual underlying set paragraph preceding Definition 7.6, pg 55 $G$, then the category in question is not a concrete category Definition 7.6 pg 55 (since the morphisms are not functions on the set $G$).

Thank you in advance
 A: Hungerford's definition of a concrete category is nonsense. A concrete category is a category $C$ equipped with a faithful functor $F : C \to \text{Set}$; to specify the data of a concrete category one must specify both what this functor $F$ does to objects and what it does to morphisms, and Hungerford's definition completely fails to specify what $F$ does to morphisms. His condition "every morphism $A \to B$ of $C$ is a function on the underlying sets" doesn't make any sense.
His counterexample is also wrong. Hungerford considers the category $BG$ with one object $\bullet$ and endomorphisms given by a group $G$, and says that if we assign this object $\bullet$ the underlying set of $G$ then we don't get a concrete category. But this is nonsense: he hasn't specified what $F$ does to morphisms! (He is also being unnecessarily confusing by identifying $\bullet$ with $G$. $\bullet$ is just a point.) And in fact there is a meaningful way to define $F$ here: you can send $\bullet$ to the underlying set $G$ and send $g \in G \cong \text{End}(\bullet)$ to the left multiplication map
$$L_g : G \ni x \mapsto gx \in G.$$
This actually does make $BG$ into a concrete category; this is essentially a restatement of Cayley's theorem, and follows abstractly from the Yoneda lemma.
