# Determining number of left cosets without Lagrange's theorem

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.

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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
...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