When is a given pair $(G,*)$ a group? I'm doing practice problems for my linear algebra class, and I don't understand how to use the group axioms to see which pairs $(G,*)$ are groups. 


*

*$G= (0,\infty)$ with $*$ given by addition

*$G= (0,\infty)$ with $*$ given by multiplication

*$G= (0,\infty)$ with $*$ given by division

*$G= (0,\infty)$ with $*$ given by $g*h = \sqrt{gh} $


I understand that I need to use the group axioms and prove the inverses etc but I don't understand what they really means and also what (G,*) means either.
 A: Some hints:


*

*$G= (0,\infty)$ with $*$ given by addition: What is the inverse of 2, for example?

*$G= (0,\infty)$ with $*$ given by multiplication: Go through this one carefully, check each axiom

*$G= (0,\infty)$ with $*$ given by division: Does $((3/4)/5) = (3/(4/5))$ ?

*$G= (0,\infty)$ with $*$ given by $g*h = \sqrt{gh} $: This is the trickiest. I'd check associativity, as we did with division.
The purpose of exercises like this is to get you used to some formalism. Abstract algebra generally, and group theory in particular, is often the first exposure newly advanced math students get to axiomatic definitions. The idea is to have you apply the abstract axioms to concrete examples in a careful fashion.
A: I'll let the other answerers address the four bullets. Just focusing on the notation "$(G, *)$", it is simply an ordered pair. We can do this because $G$ is a set and $*:G \times G \longrightarrow G$, that is, $*$ is a function (called, here, an operation). You're probably used to ordered pairs looking like "$(3,5)$", but we can construct ordered pairs from any sets as their components.
It's just a way to be clear about what set we're talking about and what operation we're talking about when discussing a particular system. Note that 
$(\mathbb{R}, +)$
is an operational system (that is, adding two real numbers always gives you a real number). In fact, it's a group. But we don't lazily say that "$\mathbb{R}$ is a group" because that sentence doesn't tell us what the operation is. To see why this is important, consider
$(\mathbb{R}, \cdot)$
which is also an operational system. However, this is not a group because it is not invertible. Hence, if we were discussing both of these operational systems, the sentence "$\mathbb{R}$ is a group" is ambiguous and potentially wrong.
Also, we shouldn't say things like "$+$ is a group" because there we haven't mentioned the base set. For example,
$(\mathbb{N}, +)$
is an operational system, but it is not a group since it doesn't even have a neutral element. So, if we were discussing $(\mathbb{N}, +)$ and $(\mathbb{R}, +)$, "$+$ is a group" would be similarly ambiguous and potentially wrong.
A: In order to see if $(G,*)$ is a group, you verify the group axioms:


*

*Closure under *: That is, if $a$ and $b$ are in $G$, is $a*b$ in $G$?

*Associativity: For $a,b,c \in G$ is it true that $(a*b)*c = a*(b*c)$?

*Existence of identity: Is there an element $e \in G$ such that for any element $a \in G$, $a * e = a$?

*Inverses: Is there an element $b \in G$ for every $a \in G$ so that $a * b = b * a = e$?
