We say $xH = yH$ iff $y^{-1}x \in H$. Writing out the definition of a coset, we have $xH := \{g\in G | g = xh, h \in H\}$, similarly $yH := \{g\in G | g = yh, h \in H\}$. If my logic is correct, $xH = yH$ implies $xh = yh$. So $y^{-1}xh = h$. I am not sure how to jump to $y^{-1}x \in H$ from this statement since we do not know if $H$ is abelian. What is the intuition behind this equality, and how can it be derived?
Intuition behind coset equality
abstract-algebraintuition
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A nice way to think about cosets is through equivalence classes.
Intuitively, if you have an equivalence relation $\sim$ on a set $X$, then this equivalence relation "lumps" the set $X$ into smaller sets, called equivalence classes. A subset $A\subset X$ is an equivalence class if there is some $a \in A$ such that $A$ is the set of all $x \in X$ satisfying $x \sim a$. A really simple example is when the set is $\mathbb{R}$ and $\sim$ is the usual equality of numbers, $=$. In this case, any equivalence class $A$ is just a singleton; $x \sim y$ for $x,y \in A$ means $x=y$. In this simple instance, $=$ "partitions" $\mathbb{R}$ into a bunch of classes (the singletons) for which each $x \in \mathbb{R}$ belongs to exactly one.
In the case of cosets, we can consider a subgroup $H\subset G$ (with identity $e$) and define a relation $\sim$ on $G$ by saying that $x\sim y$ if and only if $xy^{-1} \in H$. Indeed, this is an equivalence relation. Note the following:
We have that $xx^{-1} = e \in H$, so $x\sim x$.
If $x\sim y$, then $xy^{-1} \in H$, so $yx^{-1} = (xy^{-1})^{-1} \in H$ since $H$ is closed under inverses;
- If $x\sim y$ and $y \sim z$, then $xy^{-1} \in H$ and $yz^{-1} \in H$. As $H$ is closed under the group operation, $(xy^{-1})(yz^{-1}) = xz^{-1} \in H$. So, $x\sim z$.
These three properties are the necessary conditions for $\sim$ to be an equivalence relation. Now, we can ask, what are the corresponding equivalence classes? Well, we will see that they are sets $A\subset G$ for which $x,y\in A$ means that $xy^{-1} \in H$. That is, if we fix an $a\in A$, then elements $x\in A$ have the property that $xa^{-1} = h$ for some $h \in H$. But then $x = ha$, and since $x \in A$ was arbitrary, we can then identify $A$ with $Ha$, which is your definition of a right coset. Left cosets arise in a very similar manner by considering the equivalence relation $x\sim y$ when $x^{-1} y \in H$ (note the difference in order!).
This is a quite deep and complex topic, which certainly needs a several pages article for a decent intro into it.
Spinors (although informally, they were already in use by the end of the $19^{th}$ century) are attributed to Elie Cartan.
Intuitively (not formally), one can say that they "look" like a kind of generalization of the Euler angles: in the sense that they are used to parameterize and describe generalized rotations (in generalized spaces) in a way reminiscent to the use of the Euler angles in the parameterization of $3d$ rotations.
Cartan's initial idea involved the abstract desription of rotations of $3d$ complex vectors: We consider the complex vector space $\mathbb{C}^3$ "equipped" with the product: $$\mathbf{x}\cdot\mathbf{y}=x_1y_1+x_2y_2+x_3y_3$$ whith $\mathbf{x}=(x_1,x_2,x_3),\mathbf{y}=(y_1,y_2,y_3)\in\mathbb{C}^3$. Then we consider the set of "isotropic" (i.e.: orthogonal to themselves) vectors characterized by $$\mathbf{x}\cdot\mathbf{x}=0$$ The set of isotropic vectors of $\mathbb{C}^3$ can be shown to form a $2d$ "hypersurface" inside $\mathbb{C}^3$ and this hypersurface can be parameterized by two complex coordinates $u_0$, $u_1$: $$\begin{array}{c} u_0=\sqrt{\frac{x_1-ix_2}{2}} \\ u_1= i\sqrt{\frac{x_1+ix_2}{2}} \end{array} \ \ \ \ \textrm{or} \ \ \ \ \begin{array}{c} u_0=-\sqrt{\frac{x_1-ix_2}{2}} \\ u_1=- i\sqrt{\frac{x_1+ix_2}{2}} \end{array} $$ Cartan used the term spinor for the complex $2d$ vectors $\mathbf{u}=(u_0,u_1)$. From this, the original isotropic vector $\mathbf{x}=(x_1,x_2,x_3)$ can be easily recovered. He then proceeded to describing the rotations of $\mathbf{x}$ in terms of the rotations of $\mathbf{u}$.
For a more modern ... "skratch" on the ... "surface" of these ideas, the notes: http://ocw.mit.edu/resources/res-8-001-applied-geometric-algebra-spring-2009/lecture-notes-contents/Ch5.pdf might prove useful.
A classic -and according to my opinion, invaluable- source is the work of Claude Chevalley: "The algebraic theory of Spinors and Clifford algebras", Collected works, v.2, Springer, 1995. The classic point of view (spinors as generalized complex spaces upon which the Pauli matrices and more generally Clifford algebras act) is further analyzed. Some useful references (up to my opinion) can also be found at:
https://hal.archives-ouvertes.fr/hal-00502337/document
http://www.fuw.edu.pl/~amt/amt2.pdf
http://cds.cern.ch/record/340609/files/9712113.pdf
http://hitoshi.berkeley.edu/230A/clifford.pdf
Regarding the intuition thing about spinors. Maybe it would be useful at this point to recall that in Classical physics the description is based upon a "rigid" euclidean $3d$ background i.e. vector spaces and euclidean geometry, upon which calculus is performed and produces the quantitative prediction (which is to be tested against experiment). On the other hand, when quantum mechanics and "quantization" comes into play (in almost all elementary senses of the word quantization), the description of the states of a system is based on vectors living inside Hilbert spaces -often infinite dimensional- upon which algebras of "observables" act. The quantitative predictions are now probabilistic and consist of "spectrums" of eigenvalues of the observables.
When coming to the description of the problem of rotations, the classical physics recipe consists of using the euler angles as parameters i.e. as a kind of $3d$ coordinates leading thus to orthogonal and generalized orthogonal Lie groups. In the quantum picture, the "parameters" are now special vectors of quotient spaces of hilbert spaces, i.e. "spinors", upon which rotations, which are now for example, elements of Lie groups, Lie algebras, Pauli matrices, elements of Clifford algebras etc, act.
Best Answer
$xH=yH$ doesn't imply that $xh=yh$ necessarily. All you can deduce is that $xh_1=yh_2$ for $h_1, h_2 \in H$, ie it need not be the same element. then $xh_1=yh_2$ is equal to $y^{-1}x=h_2h_1^{-1} \in H$. Remember that $H$ is a subgroup of a group, so $h_1^{-1}$ exists, and by closure $h_2h_1^{-1}$ lies in $H$.
To show that $h$ need not be the same, look at $G=\mathbb{Z}/10\mathbb{Z}=\{0,1,2,3,4,5,6,7,8,9\}$, with $H=\{0,2,4,6,8\}$. Then $1H=\{0,2,4,6,8\}$, $2H=\{0,4,8,2,6\}$. Comparing elements, we have $1.H=2H$. For $h=4 \in H, x=1, y=2$, it doesn't hold that $1.4=2.4$, yet $xH=yH$.