What does the multiplication in cosets mean? I was getting ready to learn about order of an element, cosets, and langrange's theorem in group theory. Consequently this involves multiplying two elements of a group. After seeing several examples more abstract than multiplication of matrices, fields, and etc like with symmetries, I'm guessing multiplication in general with groups is just composition, but I've been unable to confirm that. From various other things I've already developed a mindset of multiplication (in ANY sense) and composition being the same, but I wanted to check this is also true in group theory, or concretely if a group $G$ with operation $*$ and elements $a,b,c$ that multiplication $ab$ can be defined as: 
$ab=a*b$ which follows intuitively from: 
$a*b*c=(ab)*c$ or $a*b*=(ab)*$ 
...but this does seem odd we'd introduce the new notion of multiplication here if we already have the group operator. 
Forgive me if this is a duplicate and I didn't look enough. 
EDIT: clarification, I don't know weather or not there's a general way to multiply two elements of a group (to add context, like you do in finding a coset or when raising a group element to the nth power to find it's order). If there is a general meaning of multiplication, does it basically just work like function composition, or equivelently whatever the group operator is? 
 A: I'm not sure what your question is, so instead, I'm just going to give you an example of how to multiply cosets. In this case, I will use an abelian group and the operation will be addition instead of multiplication, but the same principle applies in other scenarios.
Let's say we have $\Bbb{Z}_6$, which has the normal subgroup $N=\{0,3\}$. The following cosets exist:
$$0+N=\{0,3\}$$
$$1+N=\{1,4\}$$
$$2+N=\{2,5\}$$
$$3+N=\{3,0\}$$
$$4+N=\{4,1\}$$
$$5+N=\{5,2\}$$
Now, notice that $0+N=3+N$. This is because $3$ is part of the subgroup $N$, so $3+N=N$. Also, $4+N=1+N$ for the same reason: $4+N=1+(3+N)$, and $3+N=N$, so $4+N=1+N$. Same with $2+N=5+N$.
Now, how do we multiply (or in this case, add) these cosets? It's actually pretty simple: You just add the elements in front of the coset. For example, here's how you add $2+N$ to $4+N$:
$$[2+N]+[4+N]=(2+4)+N=6+N=0+N$$
I'll give more examples below:
$$[1+N]+[2+N]=(1+2)+N=3+N=0+N$$
$$[5+N]+[2+N]=(5+2)+N=7+N=1+N$$
$$[1+N]+[4+N]=(1+4)+N=5+N=2+N$$
Hopefully, you get the idea now. In short, if you are adding the cosets $a+N$ to $b+N$, just add $a$ and $b$ together to get $(a+b)+N$.
