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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?

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  • $\begingroup$ The operation in a group is just a rule for composing two elements of a group to get another element, I'm not sure what your actual question is? The operation on cosets is inherited from the parent group when passing to the quotient (whereby such an operation is only well-defined when your subgroup is normal). $\endgroup$ – ÍgjøgnumMeg Dec 28 '18 at 22:27
  • $\begingroup$ "I'm guessing multiplication in general with groups is just composition" -> What does this mean? Not all groups are functions (i.e. take any group with numbers, like $\Bbb Z$ under addition), so multiplication in general is not just composition. $\endgroup$ – Noble Mushtak Dec 28 '18 at 22:32
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    $\begingroup$ Every group can be thought of as a group of permutations of some set. Permutations can be composed. As I understand this is actually how group theory emerged in the first place. $\endgroup$ – lanskey Dec 28 '18 at 22:40
  • $\begingroup$ @ÍgjøgnumMeg by you don't know what the question is are you implying it's right? $\endgroup$ – Benjamin Thoburn Dec 28 '18 at 22:41
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    $\begingroup$ The title seems to imply that the question is about how the group operation is defined on cosets in quotient groups, but the body appears to address a more general question. Please clarify what is intended. $\endgroup$ – Bill Dubuque Dec 28 '18 at 23:24
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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 $(a+b)+N$.

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  • $\begingroup$ I'd upvote this now but I've reached my daily limit. $\endgroup$ – Shaun Dec 28 '18 at 23:02

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