Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I understand the standard definition of left coset, but the one I do not understand (or see why it is advantageous) is the definition that follows:

Let $H\leq G$. Then a left coset of $H$ is a nonempty set $S\subset G$ such that $x^{-1}y \in H$ for any $x,y\in S$, and for any fixed $x\in S$, the map $y\mapsto x^{-1}y $ is a surjection from $S$ to $H$.

I see how it works, but it doesn't seem to say much... What am I not seeing?

share|improve this question
add comment

1 Answer 1

up vote 5 down vote accepted

I'm not exactly sure what you mean by the canonical definition, but one standard definition of a coset is a translate $xH$ of $H$, for some $x\in G$. Another definition is that a coset is an equivalence class of the equivalence relation $\sim$ given by $x\sim y$ if $x^{-1}y\in H$. This is, in spirit, close to the definition you quote.

The equivalence class definition is useful in a couple of ways.

First, it's good to know that $\sim$ really is an equivalence relation, because that tells you immediately that the cosets partition $G$.

Second, it gives you a way of checking when two elements $x$ and $y$ represent the same coset: simply check if $x^{-1}y\in H$. This is similar to the idea of checking when two numbers $a$ and $b$ have the same remainder mod $m$: check if $m$ divides $a-b$.

The definition you quoted connects the two standard definitions -- the definition in terms of translates and the definition in terms of equivalence classes. Indeed, the fact that the transformation $y\mapsto x^{-1}y$ from $S$ to $H$ is surjective tells us that if we translate $S$ by $x^{-1}$ we get $H$ back.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.