Convention in defining a group

Question

I know that a group is a set $G$ with some operation, *. It's typical to write this as $(G, *)$. In this case, the set is separable from its operation, so it makes sense to define elements $a, b \in G$. In Artin's text, however, he defines a convention of notating a group as a set together with its operation. For example, $\mathbb{R}^{\times}$ is the group of non-zero real numbers under addition.

My question is: what is the standard for this? It seems to me that it make sense to assert that $a$ and $b$ are elements of the "group $G$" (when I mean the set $G$), though it doesn't make sense to say that $a$ and $b$ live in $\mathbb{R}^{\times}$. Further, if I define a homomorphism, it makes sense to make from a set $G$ to another set, but not from a set with some operation.

Answer 1

There is a standard abuse of notation here. In universal algebra and model theory, an algebraic structure like a group $\mathbf{G}$ is thought of very explicitly as a tuple $\mathbf{G} = (G, e, \cdot)$ comprising a carrier set or universe $G$ together with a constant $e \in G$ and an operation $(\cdot) : G \times G \to G$. In other areas of mathematics, it is common to forget the distinction between the structure $\mathbf{G}$ and the carrier set $G$.

Answer 2

When defining a group we take a set $G$ and equip it with an inner operation $*:G\times G\to G$. Writing "$a,b\in G$" remains natural as $a$ and $b$ are afterall elements of the set $G$ which happens to be equipped with a group structure aswell (keep Crostul's comment in mind: "a group is not simply a set, but it has something more").

Take $\Bbb R^\times$ under addition. We have the set $R:=\{x\in\Bbb R\mid x\neq0\}$ and we define the inner operation $+\colon\Bbb R^\times\times\Bbb R^\times\to\Bbb R^\times,~(x,y)\mapsto x+y$. Therefore $x$ and $y$ live in $\Bbb R$ as they are still non-zero real numbers but two different numbers can be "combined" (added in this case) by means of the defined operation (note, however, that $(\Bbb R^\times,+)$ is not group).

A group homomorphism $\varphi\colon(G,*)\to(H,\star)$ is a set-function $\varphi\colon G\to H$ preserving the given structure, that is $\varphi(g*h)=\varphi(g)\star\varphi(h)$. So we just assign to elements of $G$ elements of $H$ (as they are naturally elements of a set) in a way which is compatible with the given group structures.

Answer 3

It seems to me that it make sense to assert that $a$ and $b$ are elements of the ``group $G$'' (when I mean the set $G$), though it doesn't make sense to say that $a$ and $b$ live in $\mathbb{R}^{\times}$.

It actually makes sense a bit, because of how often the operation implies the set when the set is a particularly common set.

Let me be clearer and use your example. $\Bbb R^\times$ is a group of real numbers under the operation of multiplication, $\times$, right? But there's a caveat you're forgetting: for all elements of a group, there needs to be an inverse. What is the inverse of $0$ under multiplication? Obviously, nonrigorously, this would be $1/0$, which just doesn't make sense, right?

It's not that Artin is saying $\Bbb R^\times = (\Bbb R, \times)$ (i.e. establishing an equivalent notation). Or, at least, I assume so -- I haven't actually read his texts (yet). Rather, $\Bbb R^\times$ is a group all on its own, and works with a set different from $\Bbb R$. Namely, $\Bbb R^\times$ is "the multiplicative group of the real numbers," or perhaps more intuitively "the set of invertible (or nonzero) real numbers".

That is, the set $\Bbb R^\times$ is the same as $\Bbb R \setminus \{0\}$. It just also handily notates the operation, much in the same way you could say $G$ and refer to just the set, or say $G$ and refer to the group. Which is relevant is implied in the context.

More generally, when an operation $\ast$ is understood to be multiplicative - you'll understand this notion more when you deal with ring theory (in which operations may be considered additive or multiplicative) - on a set $G$, where $(G \setminus \{0\}, \ast)$ is a group, we say $G^\ast$ is a group. That is, $G^\ast = (G \setminus \{0\}, \ast)$. We would call this the multiplicative group of $G$.

(Also, in case it's not clear, $0$ in this context refers to the identity of $G$ under its additive operation, not necessarily the actual number zero itself, which may not even be in $G$. You'll often find that $(G,+)$ for an additive operation $+$ is a group and need not exclude an element. The distinction can be seen in comparing $(\Bbb R,+),(\Bbb Q,+),(\Bbb C,+)$ to, say, $(\Bbb R \setminus \{0\}, \times),(\Bbb Q \setminus \{0\},\times),(\Bbb C \setminus \{0\},\times)$ - where, this time, $0$ represents the familiar number. We remove it because it is often the "troublesome", non-invertible element under the multiplicative operation.)

We use this shorthand of $G^\ast = (G \setminus \{0\}, \ast)$ mostly out of convenience, and it tends to be commonly understood once you're used to it. It's a lot shorter than writing out the formal notation, and certainly much more so than using the definition.