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?
[Math] Let $G=\langle \mathbb{Z},+\rangle $ and $H=\{6n|n \in \mathbb{Z}\}$. Find all the distinct left and right cosets of $H$ in $G$.
finite-groupsgroup-theorynormal-subgroups
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Best Answer
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
$$y=6x+a:a,x\in\mathbb{Z}$$