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I'm having trouble understanding the motivation behind a coset. The book I'm using (A Book of Abstract Algebra) states:

Let G be a group, and H a subgroup of G. For any element a in G, the symbol

aH

denotes the set of all products ah, as a remains fixed and h ranges over H. aH is called the left coset of H in G.

It goes on to say the same about right cosets. I understand this definition (or I think I do), but what does this accomplish? Is it saying that given a subgroup H, you can generate a group G using cosets?

Thank you in advance.

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    $\begingroup$ One of the earliest cosets you have probably seen is from the context of trig equations. There a solution was often of the form $x=\alpha+n\cdot2\pi$ with $n$ an arbitrary integer. That is a coset of the subgroup $2\pi\Bbb{Z}$ within the bigger (additive) group $\Bbb{R}$. Such a coset corresponds to a unique point on the unit circle, or to a unique compass direction. So compass directions are actually cosets. Here I discuss quotient groups, they are what you get from the cosets (of a normal subgroup). Most of that applies here as well. $\endgroup$ – Jyrki Lahtonen Aug 1 '16 at 15:13
  • $\begingroup$ @JyrkiLahtonen Okay, that makes sense. I haven't delt with quotient groups yet but I know cosets are used to construct them. I guess this is one of those things I just have to take it as it is for now. $\endgroup$ – Curious Math Student Aug 1 '16 at 15:34
  • $\begingroup$ @CuriousMathStudent Are you comfortable with equivalence relations? $\endgroup$ – Alex Provost Aug 1 '16 at 17:38
  • $\begingroup$ @AlexProvost Not as much as I thought I was after reading the comments and answers. I understand what it says, but have trouble doing problems every now and then. I'll have to study them more. $\endgroup$ – Curious Math Student Aug 1 '16 at 21:51
  • $\begingroup$ I wish intro group theory books would motivate the whole enterprise with group actions instead of relegating it to a chapter at the end. The whole idea of groups is to represent symmetry, or more generally "action," in the first place. (I'm exaggerating, but only a little.) Subgroups correspond to symmetries with restrictions, homomorphisms to induced actions, and cosets to understanding the structure of orbits (via the orbit-stabilizer theorem) and the structure of homomorphic images (via the first isomorphism theorem). $\endgroup$ – arctic tern Aug 1 '16 at 23:06
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It is crucial to be comfortable with equivalence relations for this subject to make sense. Remember that an equivalence relation $\sim$ on a set $X$ is a relation satisfying three properties: reflexivity, symmetry, and transitivity. But the real reason why such a relation is interesting is that it is equivalent to a partition of $X$. A partition of $X$ is a collection of non-empty subsets of $X$ such that every element of $X$ belongs to exactly one of those subsets. The bijection between equivalence relations $\sim$ and partitions is given by $$a \sim b \iff a \text{ and } b \text{ are in the same subset}.$$ Here are some examples on finite sets:

  1. On a set $X = \{0,1,2,3\}$ of four elements, the equivalence relation generated by $0 \sim 1 \sim 2$, $0 \nsim 3$ yields the partition $\{\{0,1,2\},\{3\}\}.$
  2. On a set $X = \{0,1,2,3\}$ of four elements, the equivalence relation given by $a \sim b$ if and only if $a-b$ is a multiple of $2$ yields the partition $\{\{0,2\},\{1,3\}\}.$
  3. On a set $X = \{0,1,2,3,4,5\}$ of six elements, the equivalence relation given by $a \sim b$ if and only if $a-b$ is a multiple of $3$ yields the partition $\{\{0,3\},\{1,4\},\{2,5\}\}.$
  4. On a set $X = \{0,1,2,3,4,5\}$ of six elements, the equivalence relation given by $a \sim b$ if and only if $a-b$ is a multiple of $2$ yields the partition $\{\{0,2,4\},\{1,3,5\}\}.$

As you can see, the last three examples exhibit a symmetry in their partition, and their equivalence relations take a very specific form. This is typical of group quotients. If you have a subgroup $H$ of some group $G$, the cosets $gH$ partition $G$, and each coset has the same size. Moreover, if $G$ is finite, the order of each coset has to divide the order of $G$ (Lagrange theorem). The corresponding equivalence relation is given by $$g \sim h \iff g \text{ and } h \text{ are in the same coset} \iff gh^{-1} \in H.$$

The last three examples above are merely instances of this phenomenon for the finite groups $G = \mathbb{Z}/4,\mathbb{Z}/6,\mathbb{Z}/6$ and their subgroups $H = \{0,2\},\{0,3\},\{0,2,4\}$.

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  • $\begingroup$ Wow! Thank you for the detailed information. I stopped trying to learn abstract for now so I can focus on partitions and ERs. I can't wait to make sense of cosets, Lagrange's Theorem, quotient groups, etc. afterwards. $\endgroup$ – Curious Math Student Aug 2 '16 at 14:18
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"What does this accomplish" - Well, based on cosets you can define $G/H$ as the set of all different (left) cosets. [the same theory works with right instead]. One can prove that every two cosets are equal or disjoint, and that the union of all cosets equals $G$. Plus, the size of each coset is equal to the size of $H$. Only these basic facts can help us prove a very basic and important theorem about group theory - Lagrange's theorem, which states that if $H$ and $G$ are finite, then the size of $H$ divides the size of $G$ (and the quotient is exacly the amount of left cosets).

Also, you'll learn that if $H$ is a normal subgroup then $G/H$ is a group by itself, and understanding how $G/H$ looks can teach us a lot about $G$ itself in many different ways.

As youv'e been told in the comments, cosets are useful even without looking through the glasses of group theory. Cosets often helps us describing equivalence classes of an equivalence relation (for example, $a\sim b \Leftrightarrow a-b\in 2\pi\mathbb{Z}$).

I know it's quite superficial but you'll understand everything better as you continue learning the subject.

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  • $\begingroup$ I see. I learned about Lagrange's Theorem yesterday and normal subgroups today. But I think I'll stop there for now and backtrack to really understand cosets, partition, and equivalence relations. Thank you very much for this information. Especially that last sentence, it's very encouraging! $\endgroup$ – Curious Math Student Aug 1 '16 at 21:57
  • $\begingroup$ Your'e welcome :) If you liked my answer, (I think) you can accept it/vote for it :) $\endgroup$ – 35T41 Aug 1 '16 at 23:16
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Let me begin with a geometric example:

$\Bbb C^*$ be the group of non-zero complex numbers under multiplication

Let $H=\{x+iy\in \Bbb C:x^2+y^2=1\}$ which basically contains all the complex numbers lying on the unit circle.

Now consider the left coset $(3+4i)H$, Geometrically speaking this coset contains all points lying on the circle centered at the origin and radius $5$ (Why?)

In general the left coset $(a+ib)H$ contains all points on the circle centered at the origin and radius $\sqrt{a^2+b^2}$

Here the cosets of $H$ partition the punctured complex plane (i.e. $\Bbb C^*$) into concentric circles. Now isn't that amazing?

As seen in the above example:=

The essence of cosets lies in the fact that they partition the entire group into equivalence classes.

Moreover if the group is finite, each of the partitions will have the same number of elements, this is because of the way we define Right/Left cosets.

All the above observations leads to one of the most important result in Group Theory - The Lagrange's Theorem.

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  • $\begingroup$ Thank you for that wonderful example. It really made me have an aha moment. From the previous comments/answers, I understand that I need to have a better grasp of partitions and equivalence relations to fully understand cosets, Lagrange's Theorem, etc. So I will work on those then go back to cosets, etc. $\endgroup$ – Curious Math Student Aug 1 '16 at 22:06

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