Why category of Groups is not a subcategory/(full subcategory) of the category of Sets and notational clarifications help for Fuller and Anderson text I am trying to learn some basic category theory from various algebra text that have an introduction to category theory section.  I encounter the statement that the category of group is not a subcategory of sets.  I did some google searches, and I came across somewhat detail explanation from the book Rings and Categories of Modules by: Frank Wylie Anderson and Kent R. Fuller.  Also I know the question has already been sort of answered here @UdayReddy's response.  But I also found in another online notes that category of groups is not a full subcategory of sets from here: on page 2 right before definition 4.4 
Anderson and Fuller gave the following list of definitions concerning mappings, categories, concrete category, underlying set, the category of groups and the category of sets and subcategory:

The set of all functions from $A$ to $B$ is denoted by $B^{A}$ or by $Map(A,B)$: $B^{A}=Map(A,B)=\{f|f:A\rightarrow B\}$
The partial map $\circ$ is called the $\textit{composition},$ and the morphisms $1_{A}$ are called the $\textit{identities}$ of the category.
Let $\mathscr{C}$ be a class for each pair $A,B\in\mathscr{C}$, let $mor_{C}(A,B)$ be a set; write the elements of $mor_{C}(A,B)$ as "arrows" $f:A\rightarrow B$ for which $A$ is called the $\textit{domain}$ and $B$ the $\textit{codomain}$.
Let $C=(\mathscr{C}, mor_{C},\circ)$ be a category.  Then $C$ is $\textit{concrete}$ in case there is a function $\mu$ from $\mathscr{C}$ to the class of sets such that for each $A,B\in \mathscr{C}$ 
$mor_{C}(A,B)\subset Map(\mu(A),\mu(B)),$ $1_{A}=1_{\mu(A)},$ and such that $\circ$ is the usual composition of functions.  
Let $\mathscr{L}$ be the class of all sets; for each $A,B\in \mathscr{C}$, let $mor_{S}(A,B)=Map(A,B)$, and for each $A,B,C\in \mathscr{C}$, let $\circ:mor_{S}(B,C)\times mor_{S}(A,B)\rightarrow mor_{S}(A,C)$ be the composition of functions.  Then $S=(\mathscr{L}, mor_{S},\circ)$ is a concrete category where $\mu(A)=A$ for each $A\in \mathscr{L}$.  Call $S$ $\textit{the category of sets}$
Let $\mathscr{G}$ be the class of all group, let $mor_{G}(G,H)$ be the set of all group homomorphisms from $G$ to $H$, and again let $\circ$ be the usual composition of functions.  Then $G=(\mathscr{G}, mor_{G},\circ)$ is a concrete category, the $\textit{the category of groups}$, where $\mu(G)$ is the underlying set of $G$.
If $C=(\mathscr{C}, mor_{C},\circ)$ is a concrete category, then the set $\mu(A)$ is called the $\textit{underlying}$ set of $A\in \mathscr{C}$.
A category $D=(\mathscr{D}, mor_{D},\circ)$ is a $\textit{subcategory}$ of $C=(\mathscr{C}, mor_{C},\circ)$ provided $\mathscr{D}\subset \mathscr{C}$, $mor_{D}(A,B)\subset mor_{C}(A,B)$ for each pair $A,B \in \mathscr{D}$, $\circ$ in $D$ is the restriction of $\circ$ in $C$.  If in addition $mor_{D}(A,B)=mor_{C}(A,B)$ for each $A,B\in \mathscr{D}$, then $D$ is a $\textit{full}$ subcategory of $C.$

Anderson and Fuller then states that there are some categories which are not subsets of each other and as examples,  he compares the category of groups to the category of sets.

It is clear that the class of abelian groups is the class of objects of a full subcategory of the category of groups, and that this category has a full subcategory whose objects are the finite abelian groups.  It is a common practice in algebra to identify an object in a category with its underlying set.  Thus for example, we usually identify a group $(G, \circ)$, consisting of a set $G$ and an operation $\circ$, with its underly set $G$. Note, however, that the category of groups is not a subcategory of the category of sets, quite simply because for groups $(G,\circ)$, $(H, \circ)$ in $\mathscr{C}$ 
$mor_{G}((G,\circ),(H,\circ))\subset Map(G,H)$     (*) 
and 
$mor_{G}((G,\circ),(H,\circ))\not\subset Map((G,\circ),(H,\circ))$  (**)

I am trying to unpack the notation for (*) and (**).  Would $(G, \circ)$ and $(H,\circ)$ mean
(1) the set of homomorphic maps $w:G\rightarrow G$ with binary operation $\circ$ in both the domain and codomain and similarly for $(H, \circ)$: $z:H\rightarrow H$ with binary operation $\circ$, in $z$'s domain and codomain and so $mor_{G}((G,\circ),(H,\circ))$ then would mean the set of arrows from $w(G)$ to $z(H)$: $t:w(G)\rightarrow z(H)$ or 
(2) does $mor_{G}((G,\circ),(H,\circ))$ mean the set of arrows: $p:G\rightarrow H$, with both domain and codomain given the binary operation $\circ$
In both (1) and (2), I am understanding that the underlying set for $mor_{G}((G,\circ),(H,\circ))$ is $G=\{(G,\circ),(H,\circ)\}$ being the class of groups of homomorphism from $(G,\circ)$ to $(H,\circ)$ and does this mean it is a group homomorphism from $w(G)$ to $z(H)$ if I am interpreting as (1)
The reason I am giving two interpretations is because for the $\circ$ notation the authors use it for two different things, (1) to denote binary operation on groups, (2) to mean as a composition map. 
Furthermore, in the notation $Map((G,\circ),(H,\circ))$, the $Map(A,B)$ means $Map(A,B)$: $B^{A}=Map(A,B)=\{f|f:A\rightarrow B\}$. So in the context for the category for groups, does it mean: 
(1') with $(G,\circ)$ being interpreted as in (1) being the set of homomorphic maps $m:G\rightarrow G$ with binary operation $\circ$ in both the domain and codomain and similarly for $(H, \circ)$: $n:H\rightarrow H$ with binary operation $\circ$, in $z$'s domain and codomain, so $Map((G,\circ),(H,\circ))$ would mean $Map((G,\circ),(H,\circ))=\{q|q:m(G)\rightarrow n(H)\}$
(2') $Map((G,\circ),(H,\circ))=\{f|f:(G,\circ)\rightarrow (H,\circ)\}$ with both domain and codomain given the binary operation $\circ$
Given my lack of clear interpretations of the notations, I am not clear why (*) and (**) holds.  Also, as to why the category of groups is not a subcategory of the category of sets, beside not understanding exactly why that is, does the statenent imply the two are not considered to be full subcategory of each other?  Thank you in advance.
 A: Here's a good way to figure out what $ \circ $ means. In this context, If it appears as $(G, \circ)$ or $(H, \circ )$ then it's the group multiplication. Otherwise it's probably composition in a category.
To write it out: If you have $(G, \circ)$ then here $\circ$ is a set-map $G \times G \rightarrow G$ with certain conditions.
$mor_G((G,∘),(H,∘))$ is the set of group homomorphisms between the two groups. There's bad notation being used here since $G$ refers to the category of groups (which is better denoted by something like $\mathsf{Grp}$) and to a certain group.
You don't need to look at homomorphic maps $G \rightarrow G$ or $H \rightarrow H$ here.
(*) Here's the crux of the argument: Any homomorphism of groups $f: (G,\circ) \rightarrow (H, \circ)$ is a function/set-map $f: G \rightarrow H$ (with additional properties) where $G$ and $H$ are the sets underlying the respective groups. So, $mor_G((G,∘),(H,∘))\subset Map(G,H)$.
(**) $Map((G,∘),(H,∘))$ is a different beast entirely, since it is the set of (set-)functions from $(G,\circ)$ to $(H, \circ)$. With very mild abuse of notation, note that as a set, $(G,\circ) = (G, X)$ where $X$ is the appropriate subset of $(G \times G) \times G$ which represents $\circ$. Depending on your definition of an ordered pair, $(G,X)$ looks something like $ \{G, \{G, X \} \}$.
Ultimately an element of $mor_G((G,∘),(H,∘))$ is a set-map $G \rightarrow H$,
and the elements of $Map((G,∘),(H,∘))$  are not of this type.
