A problem on group theory: "Multiplication defined by composition" I recently encountered this question:

Suppose $G$ is the set of all bijective functions from $\mathbb{Z}$ to $\mathbb{Z}$ with multiplication defined by
composition, i.e., $f\cdot g = f \circ g$. Prove that, $(G, \circ )$ is a group but not an abelian group.

I'm having some trouble understanding this question.
First, what exactly does "multiplication defined by composition" mean? Can you please give an example?
Second, can you please show me how to prove this question?
Thank you for taking the time to reply to this.
 A: A group $G$ is endowed with a binary operation $*$ that maps $x,y \in G$ to $x*y \in G$.
On your case, $G$ is the set of bijective maps from $\mathbb Z$ to $\mathbb Z$. And $*$ the composition of maps.
To verify that $(\mathbb Z, \circ)$ is a group, just verify that it satisfies the axioms of a group.
To prove that it is not abelian, find two maps such $f,g$ that $f \circ g \neq g \circ f$.
A: By multiplication, in this case, we mean function composition.
So, you need to prove that the four conditions for a group are satisfied.  Namely, closure under addition, existence of inverses, existence of an identity element, and associativity (not necessarily in that order).
Once you've done that, you know you have a group.  But if you can exhibit a pair of maps $f$ and $g$ such that $f\circ g\ne g\circ f$, you will have shown it is not abelian.  For this last part, how about defining $f:\Bbb Z\to\Bbb Z$ by $f(0)=1, f(1)=0$, and $f(x)=x$ otherwise.  And define $g:\Bbb Z\to\Bbb Z$ by $g(x)=x+1$.  Then  $(f\circ g )(0)=0$, but $(g\circ f)(0)=2$.  So they don't agree.
A: A group $G$ is a set with an operation $\circ$ that has certain properties.
For example the set $\mathbb{Z}$ is a group with the operation $+$.
Most groups are written "multiplicativley", so we do not have to write this much. So one could say the group operation is multiplication, and we would write for example $xy$ instead of $x+y$.
In your example we have as set the set of bijective functions from $\mathbb{Z}\to\mathbb{Z}$.
So we need an operation that makes sense for functions. And the only operation that comes to mind is the composition $\circ$ of functions.
We just call this group multiplication. It is just a name and has nothing to do with the multiplication you know otherwise.
For example the set $\mathbb{Z}$ would not be a group when we take the multiplication $\cdot$ as operation, as a group must have inverse elements, so for $x\in\mathbb{Z}$ we must have $y\in\mathbb{Z}$ with $x\cdot y=y\cdot x=1$.
For $2\in\mathbb{Z}$ there would be not such inverse, as $\frac12\notin\mathbb{Z}$.
With the group operation $+$ however everything works.
So when you work with groups, you have to be aware what the operation is. Especially because we denote a group mostly just by the set and not talking about the operation.
To be extra specific one would say let $(G,\circ)$ be a group, but most of the time we just say let $G$ be a group. But then the operation is obvious.
For someone starting out in math this is confusing, because they would not ask themself the question "what is the operation here?" and get confused.
