Let $G=\langle \mathbb{Z},+\rangle $ and $H=\{6n|n \in \mathbb{Z}\}$. Find all the distinct left and right cosets of $H$ in $G$.

Question

I have an exercise where I am supposed to find the left and right cosets. But how do I generate the cosets? As I have understood it you are supposed to pick a number that is not in the set $H$ and multiply it with every number in $H$. But this does not exactly give the right answer. I would really appreciate it if someone gave an easy to understand explanation of how to generate the left and right cosets. Also is this group a normal subgroup?

Answer 1

To get you started:

• The group operation is $+$, so you want to take an element in $G$ and add it to all of the elements in $H$. This will be a left or right coset, depending on what side you add from.

• The addition operation in $\mathbb{Z}$ is abelian, so the left and right cosets will be the same. In particular, $g+H = H+g \forall g \in G$, so $H$ is a normal subgroup of $G$.

Answer 2

Let $a\in G$. In other words, $a$ is an integer.

Since the group is defined by addition, the left coset $aH$ is simply the set $\{a+6n\}$ with $a,n \in \mathbb{Z}$

In particular, addition is commutative so the group operation in $\mathbb{Z}$ is abelian, leading us to reasonably claim that the left and right cosets are equal. (You can verify this)

Geometrically, you can interpret this set as the set of all

$$y=6x+a:a,x\in\mathbb{Z}$$

Answer 3

If $H$ is a subgroup of $G$, then $gH$ is called a left coset of $H$ in $G$. Here, $gH$ is defined to be the set $\{gh: h \in H\}$. The set of all left cosets of $H$ in $G$ is defined to be the set $\{gH: g \in G\}$.

In your case the group $G$ is the set of integers under addition. The subgroup $H$ is the set $6 \mathbb{Z} = \{6n: n \in \mathbb{Z} \}$ of all integers which are multiples of $6$. Each coset of $H$ in $G$ is of the form $gH$ for some $g \in G$. If you take $g=2$, then $gH$ is the coset $2+6 \mathbb{Z}$. Observe that the group operation is addition, so that $gH = \{gh: h \in H\}$ becomes $\{2+k: k \in 6\mathbb{Z} \}$. This coset is the set of all integers which give a remainder of 2 when divided by 6.

There are a total of 6 left cosets of $6 \mathbb{Z}$ in $\mathbb{Z}$, and they are of the form $i+6 \mathbb{Z}$ for $i=0,1,\ldots,5$. Each left coset $i+6 \mathbb{Z} = \{i+6n: n \in \mathbb{Z} \} $ is equal to the right coset $6 \mathbb{Z}+i$ because the group operation is commutative. So, the set of all right cosets is exactly the set of all left cosets.

Since the group $G$ is abelian, every subgroup $H$ is normal in $G$, ie $gH=Hg$ for each $g \in G$. To prove that $H$ is normal, we only require that $gH=Hg$ for each $g \in G$. In this example, because the group operation is commutative, the group satisfies the even stronger property that $gh=hg$ for each $g \in G$.