Why is this not a group? It's been a while since I took abstract algebra but I'm wondering what is wrong with my reasoning here
A group $(G, \circ)$ is defined as a tuple that consists of a set $G$ along with an operation $\circ: G \times G \to G$ such that the following conditions are satisfied

*

*(Closure) For all $a,b \in G$, $a \circ b \in G$

*(Associativity) For all $a,b,c \in G$, $(a \circ b) \circ c = a \circ (b \circ c)$

*(Identity) There is an element $1_G \in G$ such that $a \circ 1_G = 1_G \circ a = a$ for all $a \in G$

*(Inverse) For all $a \in G$, there is an element $a' \in G$ such that $a \circ a' = a' \circ a = 1_G$
Now, as far as I know, the uniqueness of the identity and inverse are not axioms of the group definition.
So now consider the following:
Let $G = \{1_G, a, b, c, d\}$ and define the operation $\circ$ such that $a \circ b = 1_G$ for all $a,b$

My question now is whether $(G,\circ)$ is a group. Simple algebra can show that in fact $a = b = c = d = 1_G$ which makes sense since $(\{1\}, \cdot)$ is a group where $1 \cdot 1 = 1$.

But the fact that $a = b = c = d = 1_G$ doesn't make sense to me since $a,b,c,d$ could be anything. At its core, $G$ is a set and I might as well have written $G = \{1_G, \text{horse}, \text{monkey}, \text{dog}, \text{cat}\}$ with the earlier specified operation.
If $G$ is looked at as just a set with no other structure, then clearly $\text{monkey} \neq \text{horse}$ but somehow now if you combine it with an operation, then suddenly $\text{monkey} = \text{horse}$? Or is something else going on here that I seem to have forgotten...
 A: "...since $a,b,c,d$ could be anything." What you just concluded is that this claim is false if $(G,\circ)$ is indeed a group (that is, if $(G,\circ)$ is a group, then we must have $a = b = c = d = 1_G$ necessarily, no other choices). If it isn't then of course it can be anything else, like you said horse, monkey etc.
In other words, should you insist that $a,b,c,d$ are distinct objects, then $(G,\circ)$ can't be a group.
A: If $G$ is a group then for all $g\in G$ we have that $gG=G=Gg$. However, in your example you have that $gG=\{1_G\}$, so this is not a group (it is a semigroup though). You should think of the elements in your set/group as constants, and so they are implicitly non-equal and cannot be "anything".
The property that "for all $g\in G$, $gG=G=Gg$" actually almost characterises groups. You additionally need $G$ to be non-empty. So a  group is the same as a nonempty set with an associative binary operation $\cdot :G\times G\rightarrow G$ called multiplication such that for any $g\in G$, the operations of left and right multiplication by g are bijections. This characterisation leads to the idea of the "empty group", and there is a recent $n$-category cafe post about this here (from which I stole the above definition).
