What is the operation through which a Coset is created?

Question

Most books initially explain Cosets in terms of $\mathbb{Z}$ & then use the subgroup $n\mathbb{Z}$ to partition into $n$ Cosets.

1. Is the group $\mathbb{Z}$ used here the additive $\mathbb{Z}^+$ or is it the multiplicative $\mathbb{Z}^*$? I think it's $\mathbb{Z}^*$ but want to be sure.

2. What is the operation $n\mathbb{Z}$ here? Is it the multiplication of each element of $\mathbb{Z}$ by $n$ or adding the element $n$ times?

3. What is the abstraction of the operation $n\mathbb{Z}$ in groups other $\mathbb{Z}$ or other number groups? If there is some other kind of group which doesn't involve numbers, how do you do the operation corresponding to "$n\mathbb{Z}$" in that group to create cosets?

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UPDATE to my question:

I am using Contemporary Abstract Algebra by Gallian. He has a list of properties of Cosets.

One of the properties is

• For a group $G$ & a subgroup $H$, $aH = H$ if $a \in H$ - i.e. you generate a new coset only if $a$ doesn't belong to $H$.

The proof of this is that if $a \in H$, then a * h for each $h \in H$ generates an element of $H$ because of closure in $H$. However if H were an additive group then a*h is not an operation of H & hence why would there be closure. Hence I am assuming that H is a multiplicative group.

Answer 1

Most books initially explain Cosets in terms of $\mathbb{Z}$ & then use the subgroup $n\mathbb{Z}$ to partition into $n$ Cosets.

1. Is the group $\mathbb{Z}$ used here the additive $\mathbb{Z}^+$ or is it the multiplicative $\mathbb{Z}^*$? I think it's $\mathbb{Z}^*$ but want to be sure.

The group is $\mathbb{Z}^+$, as $\mathbb{Z}$ under multiplication does not form a group. (Hint: What should the identity for $\mathbb{Z}^*$ be? Then what is the inverse of $2$, say? [Alternative issue: the element $0$.])

2. What is the operation $n\mathbb{Z}$ here? Is it the multiplication of each element of $\mathbb{Z}$ by $n$ or adding the element $n$ times?

This is the set $\{nx\mid x\in\mathbb{Z}\}$, i.e. all multiples of the integer $n$. It is a normal subgroup of $\mathbb{Z}$, which is important because we need normal subgroups to form quotient groups.

3. What is the abstraction of the operation $n\mathbb{Z}$ in groups other $\mathbb{Z}$ or other number groups? If there is some other kind of group which doesn't involve numbers, how do you do the operation corresponding to "$n\mathbb{Z}$" in that group to create cosets?

The $n\mathbb{Z}$ sets are important here because they are precisely the subgroups of $\mathbb{Z}$. That is all that is going on.

In the general situation, any subgroup $H$ of a (multiplicatively-written)* group $G$ has cosets $$G/H=\{gH\mid g\in G\}.$$ (We actually talk about "left" and "right" cosets, $gH$ and $Hg$, but lets ignore that here.) Cosets are especially important as if the subgroup $H$ is a "normal" subgroup, so has the additional property that $g^{-1}Hg=H$ for all $g\in G$, then $G/H$ forms a group. This is the case with $\mathbb{Z}/n\mathbb{Z}$, as every subgroup of $\mathbb{Z}$ is normal (because it is abelian).

*Groups are typically written with a multiplicative operation. However, $\mathbb{Z}$ is written additively. Here, your cosets are $\mathbb{Z}/n\mathbb{Z}=\{a+n\mathbb{Z}\mid a\in\mathbb{Z}\}$.

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Response to the update: As your group is additively-written, "closure" becomes $a+h\in H$. The operation $∗$, which we call "multiplication", denotes an abstract group operation, but when dealing with concrete groups you need to replace it with the relevant operation. Addition here, function composition for permutation groups, etc.

So we can translate Gallian to use additive notation:

• For a group $G$ & a subgroup $H$, $a+H = H$ if $a \in H$ - i.e. you generate a new coset only if $a$ doesn't belong to $H$.

The proof of this is that if $a \in H$, then $a+h$ for each $h \in H$ generates an element of $H$ because of closure in $H$.