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I am trying to understand left/right cosets in group theory.

Here is the example in my text:

Let $G = \lbrace 1, a, b, c, d ,e \rbrace$

Lets define the group operation $.$ by the following table, where the entry at row $x$ and column $y$ gives $x.y$

ex. $d.e = b$

 1 a b c d e
1 1 a b c d e
a a b 1 d e c
b b 1 a e c d
c c e d 1 b a
d d c e a 1 b
e e d c b a 1

This is no problem i understand this. But then we get the left and right cosets.

Let $G$ be a group and let $H \leq G$. A left coset of $H$ in $G$ ($G / H$) is a set of the form $gH = \lbrace gh : h \in H \rbrace$ for some $g \in G$. A right coset of $H$ in $G$ ($H$ \ $G$) is a set of the form $Hg = \lbrace hg : h \in H \rbrace$ for some $g \in G$.

Here are some examples that i am trying to figure out how they are generated. I guess i just do not understand the theory fully. I would like a little help explaining and possible a few more examples.

$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, a, b \rbrace = \lbrace \lbrace1, a, b\rbrace , \lbrace c, d, e \rbrace\rbrace$
$\lbrace 1, a, b \rbrace$ \ $\lbrace 1, a, b, c, d, e\rbrace = \lbrace \lbrace 1, a, b \rbrace, \lbrace c, d, e \rbrace \rbrace$
$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, c \rbrace = \lbrace \lbrace 1, c \rbrace, \lbrace a, d \rbrace, \lbrace b, e \rbrace \rbrace$

A few more examples which i want to figure out are:
$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, e \rbrace = ?$
$\lbrace 1, d \rbrace$ \ $\lbrace 1, a, b, c, d, e \rbrace = ?$

Thanks!

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What textbook are you using gives examples of groups by giving their multiplication table?! –  Mariano Suárez-Alvarez Oct 9 '10 at 22:26
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The book i am using is for a CS course on Combinatorial Algorithms, "Classification Algorithms for Codes and Designs" –  gprime Oct 9 '10 at 23:02
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5 Answers 5

up vote 11 down vote accepted

If your table is correct, it should be the symmetric group in three letters, with a and b the 3-cycles, and c, d, and e the transpositions.

Now. Your first equations is, alas, nonsense. The set that contains $1$, $a$, $b$, $c$, $d$, and $e$ is certainly not equal to the set whose elements are the sets $\{1,a,b\}$ and $\{c,d,e\}$. You are writing nonsense. The rest are a bit better because you are looking at cosets. Fixed in problem.

$G$ has six different subgroups: the trivial subgroup $\{1\}$; the whole subgroup $G$; one subgroup of order three, $H=\{1,a,b\}$, and three subgroups of order two: $K_1=\{1,c\}$, $K_2=\{1,d\}$, and $K_3=\{1,e\}$.

What are the left cosets of $H$ in $G$? They are the sets $1H$, $aH$, $bH$, $cH$, $dH$, and $eH$. As it happens, $1H=aH=bH = H$, and $cH=dH=eH=\{c,d,e\}$. You can verify it exlicitly; for instance, $$aH = \{ ah : h \in H\} = \{a1, aa, ab\} = \{a, b, 1\} = H$$ and $$cH = \{ ch: h \in H\} = \{c1, ca, cb\} = \{c, e, d\}.$$

The right costs of $H$ are the same as the left cosets.

Now, what are the left cosets of $K_3=\{1,e\}$ in $G$? They are the sets $1K_3$, $aK_3$, $bK_3$, $cK_3$, $dK_3$, and $eK_3$. They are: $$\begin{array}{rcl} 1K_3 & = & \{1k : k \in K_3\} = \{11, 1e\} = \{1,e\} = K_3\\ aK_3 & = & \{ak : k \in K_3\} = \{a1, ae\} = \{a, c\}.\\ bK_3 & = & \{bk : k \in K_3\} = \{b1, be\} = \{b, d\}.\\ cK_3 & = & \{ck : k \in K_3\} = \{c1, ce\} = \{c, a\} = aK_3.\\ dK_3 & = & \{dk : k \in K_3\} = \{d1, de\} = \{d, b\} = bK_3.\\ eK_3 & = & \{ek : k \in K_3\} = \{ei, ee\} = \{e, 1\} = K_3. \end{array}$$ So there are three distinct left cosets, and they are $\{1,e\}$, $\{a,c\}$, and $\{b,d\}$.

What are the right cosets of $\{1,d\}$ in $G$? They are $K_21$, $K_2a$, $K_2b$, $K_2c$, $K_2d$, and $K_2e$. Let us be a bit smarter this time, rather than computing them directly: we know that any two distinct cosets are disjoint, and that the right coset of $x$ contains $x$. So $K_21$ contains both $1$ and $d$, hence must be equal to the right coset of $d$, $K_2d$; that is, $K_21=K_2d = \{1,d\}$. The right coset of $a$ contains $a$ and contains $da = c$, hence must be equal to the right coset of $c$, $K_2c$; indeed it is, as $dc = a$; so $K_2a=K_2c=\{a,c\}$. And the right coset of $b$, $K_2b$, contains $b$ and $db=e$, hence must equal the right coset of $e$, $K_2e$, which it does (since $de=b$ and $1e=1$). So $K_2b=K_2e=\{b,e\}$. So the three right cosets of $K_2$ in $G$ are $\{1,d\}$, $\{a,c\}$, and $\{b,e\}$.

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sorry i fixed the first equation –  gprime Oct 9 '10 at 21:39
    
wow, this helped out a lot, thanks!!! –  gprime Oct 9 '10 at 21:43
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The fundamental idea of a coset is that of taking a subgroup and "translating" it to fill up $G$. A coset is just one such "translate." No coset but $H$ itself is a subgroup of $G$, since they don't contain $1$, but there is a lot of ground that can be covered by treating them like elements of a group or set instead. Since the cardinality of the cosets is constant, they represent ways of dividing up $G$, and you get nice formulas like $|G|=|H|[H:G]$, where $[H:G]$ is called the index of $H$ in $G$ and is equal to the number of cosets.

In my opinion, the biggest use of cosets is to set up the idea of a quotient group. You may have seen the general idea -- formally dividing one mathematical structure by a substructure -- in other fields, like topology. What you do in almost every case is add, to the laws defining your mathematical structure, another law that sets everything in the substructure equal to a point, or the identity, or zero. Then you try to make everything else work.

In group theory, in particular, the set $G/H$ (or $H\backslash G$) always has an induced group structure when $H$ is a normal subgroup. And in fact, it's easy most of the time to just think of its elements as abstract and indivisible group elements. But sometimes, especially when you're proving that this works, it's useful to think of them as equivalence classes instead, where the equivalence relation is of the form $a\sim b$ when $a=bh$ for some $h\in H$. Of course, such an equivalence class is just the coset $bH$.

I'll give one nice example with abelian groups before moving on.

The quotient group $\mathbb{R}/\mathbb{Z}$ is the set of cosets of the form $x+\mathbb{Z}$, with $(x+\mathbb{Z})+(y+\mathbb{Z})=(x+y)+\mathbb{Z}$, which is just the induced addition. But you can also think of the elements of the group as numbers of the form $0\le x<1$, with addition being done "modulo $\mathbb{Z}$": that is, $x+y$ in $\mathbb{R}/\mathbb{Z}$ is equal to the fractional part of $x+y$ in $\mathbb{R}$. So all we really have is a circle with an addition operation. One place where this pops up is the group formed by numbers $e^{i\theta}$, the unit circle in $\mathbb{C}$, under multiplication.

(Another, weirder example is $\mathbb{R}/\mathbb{Q}$. I don't know much about it in terms of group theory, but it gives rise to a standard example of a non-measurable set in analysis.)

Another use of cosets will pop up when you study group actions. I think this is just called the coset action. Basically, given a subgroup $H$, $G$ acts on $G/H$ by multiplication: $(g,kH)\mapsto gkH$. Surprisingly, this action works differently than $G$'s normal multiplication on itself. I don't know any good examples of this.

Now let's look at the coset questions you mentioned.

First, the line $\lbrace 1,a,b,c,d,e\rbrace=\lbrace\lbrace 1,a,b\rbrace,\lbrace c,d,e\rbrace\rbrace$ looks like a typo or something.

Now let's look at the right cosets of $\lbrace 1,a,b\rbrace$ in $G$. If you look at the top left corner of the multiplication table, you see that right multiplication by $a$ or $b$ just gives you the same set $\lbrace 1,a,b\rbrace$, though its elements are permuted. Multiplying by $c,d,$ or $e$ gives something different. We don't have to check that these all give the same coset, since the sizes of the cosets are identical.

In the second example, you can look at the two columns labeled $1$ and $c$. Then just scan down the rows: $a\mapsto\lbrace a,d\rbrace,b\mapsto\lbrace b,e\rbrace,c\mapsto\lbrace c,1\rbrace$, and so on. These are all cosets, and identifying three distinct ones is easy. They also show you something important: if, for example, $aH=\lbrace a,d\rbrace$, you know for a fact that $dH$ is the same set, simply because $H$ contains $1$. Thus, the elements of a coset always give the same coset when multiplied by the subgroup.

Now let's do the same thing for the left cosets of $\lbrace 1,e\rbrace$. Again, ignore everything but the $1$ and $e$ columns, and scan down. We get $aH=\lbrace a,c\rbrace =cH,bH=\lbrace b,d\rbrace =dH$. And, of course, $eH=H$ since $e\in H$.

The right cosets of $\lbrace 1,d\rbrace$ will be similar, but you'll be using the rows $1$ and $d$ instead of columns.

Good luck, and I hope this was at least somewhat helpful!

Also, what book are you using? I learned all this from Artin, but it's possible that you might be getting a different perspective if you have a different book.

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Thanks, yes this is helpful. The book i am using is for a CS course on Combinatorial Algorithms, "Classification Algorithms for Codes and Designs" –  gprime Oct 9 '10 at 21:45
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Some general notes: the notation G/H is used to denote the set of left cosets of H in G. Note that this is a set whose elements are also sets (in particular, they are the left cosets of H). Similarly, the notation H\G is used to denote the set of right cosets of H in G; again, this is a set whose elements are sets.

So, let's look at the first three equations. Unfortunately, the first one makes no sense (OP has since edited the first equation), so let's start with the second one.

Notationally, as per what I wrote above, {1,a,b} \ {1,a,b,c,d,e} denotes the set of right cosets of {1,a,b}. To actually compute this, you take each element of H={1,a,b} and multiply on the right by each element of G={1,a,b,c,d,e}. So for example:

H*1 = {1,a,b}*1 = {1*1, a*1, b*1} = {1,a,b} = H

But that's too easy. Try the element a:

H*a = {1,a,b}*a = {1*a, a*a, b*a} = {a,b,1} = H, where we're using the multiplication table given to figure out each of these products.

Proceed similarly for each of the remaining elements of G (which are b,c,d,e). That is, compute the cosets H*b, H*c, H*d, H*e in exactly the same way we did above. You should find that every coset is either H or {c,d,e}. Thus, H\G = { H , {c,d,e} }.

For the third equation, you want to compute the left cosets of the subset {1,c}. You do this exactly as I've done above, except remember to multiply on the left.

For the two examples you'd like to see worked out, we again proceed exactly as above. The only difference is that the first example is left cosets of {1,e} while the second example is right cosets of {1,d}.

I highly recommend working through these examples on your own. Working with cosets can be a little goofy at first, but doing the legwork on your own is really invaluable.

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thanks yeah i worked a few out and understand them now. –  gprime Oct 10 '10 at 17:59
    
Interesting -- this page has the notational convention for the sets of left and right cosets the other way around. –  joriki Nov 13 '12 at 7:32
    
@joriki: Well, I was simply using the same notation as OP. –  Bey Nov 14 '12 at 5:19
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gprime, thinking about cosets in the following context may help you out. Leg $G$ be $\mathbb{R}^3$, the group operation will be vector addition. Suppose you want to know the left cosets of $H$, where $H$ is a straight line through the origin, which as you know, is a subspace. You look at the sets "$x+H$". Fix such an $x$, and see what happens when you add $x+h_1$, $x+h_2$, etc. where $h_1, h_2, ...$ are elements of $H$. You get a straight line that is parallel to $H$, but no longer passes through the origin. Which point does it surely pass through? For one, $x$, because, being $H$ a subspace of $\mathbb{R}^3$, $x+0$ is one of the elements of $x+H$. If you change the $x$ for $y$, another point not in the same straight line that is $x+H$, you'll obtain another straight line parallel to $H$. The left cosets (which in this case are the same as the right cosets, because vector addition is commutative) are thus all straight lines parallel to $H$. Since in this case the left and right cosets are the same, all these parallel straight lines form a group themselves, with the group operation defined simply as this: to add two straight lines, just take one point from on one of them, another point on the other, and add them, and all points that you obtain doing this will be a new straight line parallel to $H$. This new group is called the quotient group. As your textbook says, however not all left cosets will be right cosets, in which case there's no defining a quotient group. Well, I hope I didn't make it worse, but that particular example was helpful for me.

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Cosets arise when you want to model the idea that certain elements of a group are effectively equal.

To see this, rather that looking at one coset at a time it is best to think of all possible cosets of a subgroup. Then you will find that cosets partition a group $G$ into equivalence classes such that two elements of a class differ by an element of the subgroup $H$.

For example sometimes in number theory we do not want to distinguish two numbers that differ by a multiple of a given number $n$. Say $n=4$. Then let $G$ be the set of integers under addition and $H$ be the set of multiples of $4$. The cosets of $H$ are

$$H\cdot 0=\{\ldots,-4,0,4,8,\ldots\}$$ $$H\cdot 1=\{\ldots,-3,1,5,9,\ldots\}$$ $$H\cdot 2=\{\ldots,-2,2,6,10,\ldots\}$$ $$H\cdot 3=\{\ldots,-1,3,7,11,\ldots\}$$ $$H\cdot 4=\{\ldots,0,4,8,12,\ldots\}$$ $$H\cdot 5=\{\ldots,1,5,9,13,\ldots\}$$

Note that $H\cdot 0 =H\cdot 4$ and $H\cdot 1 =H\cdot 5$. In fact there are only $4$ distinct cosets, each corresponding to one congruence class of integers modulo $4$. If we care only about the remainder of an integer after division by $4$ then all the elements in a coset are equivalent and we can think of them as a single entity.

To take another example, let $G$ be the group possible rotations of a point on a unit circle and we care only about where a point ends up on the circle after the rotation. Then let $H$ be rotations that are a multiple of $2\pi$. The cosets of $H$ will now be rotations which differ by multiples of $2\pi$ and which therefore have the same effect on the final position of the point rotated.

In general, two elements $g_1$ and $g_2$ of $G$ are defined to be equivalent ($g_1 \equiv g_2$) if $g_1 \cdot g_2^{-1} \in H$.

You can prove the following

  1. For all $g$ in $G$, $g \equiv g$ since the identity element belongs to $H$ as it is a subgroup.
  2. If $g_1 \equiv g_2$ the $g_2 \equiv g_1$ since if an element belongs to $H$ then so does its inverse.
  3. If $g_1 \equiv g_2$ and $g_2 \equiv g_3$ then $g_1 \equiv g_3$ since $H$ is closed under multiplication.

This show that $\equiv$ is a genuine equivalence relation. We define the equivalence class of an element $x$ of $G$ as $$[x]=\{y \in G\mid x \equiv y\}$$

You can show that

  1. $x \in [x]$
  2. For any $x,y \in G$, either $[x] \cap [y] =\emptyset$ or $[x]=[y]$.

So the equivalence classes are either equal or disjoint and they cover $G$.

Finally, back to cosets. The equivalence classes we have defined above are the same as the right cosets. If $y \equiv x$ then $yx^{-1}=h$ for some $h \in H$ or $y=hx$. So $[x]=Hx$. If we had instead defined our equivalence relation by the condition $x^{-1}y \in H$ then we would have got the left cosets.

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