MHB What are Left Cosets and How Do You Find Them?

[karush]

Let $H=\{0;\pm 3;\pm 6;\pm 9;\cdots\}$ Find all the left cosets of $H \textit{ in } \Bbb{Z}$ok I can only see that
From $\textit{H}$ we have $\textit{H}=3 \Bbb{Z}$ thus we cosets of
$1+3\Bbb{Z},\quad 2+3\Bbb{Z} \cdots $ didn't know what the "left cosets" meantthere must be more that could be said about this
so more info welcome

 

[GJA]

Hi karush,

didn't know what the "left cosets" meant

Let $H$ be a subgroup of $G$. The left cosets of $H$ are all sets of the form $gH$, where $g\in G$ and $gH=\{gh:~h\in H\}$ by definition. Note that even though $H$ is a subgroup of $G$, $gH$ is not necessarily a subgroup (which is why the phrasing "sets of the form..." was just used); e.g, the left coset $1+3\mathbb{Z}$ of your original post is only a subset of $\mathbb{Z}$, even though $3\mathbb{Z}$ is a subgroup of $\mathbb{Z}$.

With this in mind, when you list all the possible cosets you must let $g$ run through all the elements of the group $G$. In your example this means looking at $k+3\mathbb{Z}$ for all $k\in\mathbb{Z}$. Upon doing so you should find that two cosets $k_{1}+3\mathbb{Z}=k_{2}+3\mathbb{Z}$ if and only if $k_{1}\equiv k_{2}$ (mod 3). Hence, there are only 3 left cosets of $3\mathbb{Z}:$ $3\mathbb{Z}, 1+3\mathbb{Z}$, and $2+3\mathbb{Z}$. In essence this proves a well-known result from abstract algebra that $\mathbb{Z}_{n}\cong \mathbb{Z}/n\mathbb{Z}.$

there must be more that could be said about this

There certainly is. All of this talk about cosets is leading to the notion of normal subgroups and quotient/factor groups. For example, in the result I quoted above, $\mathbb{Z}_{n}\cong \mathbb{Z}/n\mathbb{Z}$, $\mathbb{Z}/n\mathbb{Z}$ is a quotient/factor group. As you will soon learn, all subgroups of an abelian/commutative group, which $\mathbb{Z}$ is, are necessarily normal. Quotient groups are an important tool, so it is good that you are working hard to understand left cosets.