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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My book (Michael Artin's Algebra does not explain how left cosets were obtained. My difficulty understanding is from the following passage:

For example, in the symmetric group $S_{3}$, (the permutation for (123) is $x$, and the one for (12) is $y$, and so $S_{3} = \{1, x, x^{2}, y, xy, x^{2}y\}$,) the element $y$ generates a cyclic subgroup $H = \langle y\rangle$ of order 2. Then there are three left cosets of $H$ in $G$:

$ H = \{1,y\} = yH$

$ xH = \{x,xy\} = xyH$

$x^{2}H =\{x^{2},x^{2}y\} = x^{2}H$.

My question is, why are the elements in the 3 cosets grouped in the way they are?

Thanks.

share|cite|improve this question
2  
Well, I urge you to stop right here: Read (Reread) the definition of left cosets. They are translates of a subgroup $H$ in $G$. So, I translate the subgroup $H=\{1,y\}$ by an element $a$ -- then, I have $aH=\{a \cdot 1, a \cdot y\}$. So, if $a \in H$, for instance, $a=y$, then, I get back the same subgroup. This explains: $H=yH$. That is, $aH=H$ if $a \in H$. – user21436 Aug 26 '12 at 2:40
2  
...and after you've done what wentaway adviced you, think that how we group elements within some set is, usually, pretty unimportant. – DonAntonio Aug 26 '12 at 2:44

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.