# Question about wiki artical on Quotient Groups

1. Apr 30, 2012

### Diffy

Hi I am trying to learn about quotient groups to fill the gaps on things I didn't quite understand from undergrad. Anyway I have a question regarding this:

Can someone please explain how { 0, 2 }+{ 1, 3 }={ 1, 3 } in Z4/{ 0, 2 }?

I would think since 0 + 1 = 1 and 2 + 3 = 1 under mod 4 addition we would not get such a result.

Thanks,

-Diffy

Last edited: Apr 30, 2012
2. Apr 30, 2012

### Stephen Tashi

Did you mean: "since 0 +1 = 1 and 2 + 3 = 1 under mod 4 addition"?

Maybe you should think more abstractly. If you grant that {0,2} {1,3} is a group with 2 elements, then one of them has to be the identity, correct?

A concrete way of looking at it would be that that the operation you asked about amounts to ( (X + Y)(mod 4) ) mod 2 except that the equivalence classes for mod 2 aren't represented by the two elments {0}, {1}. They are represented by two sets of elements {0,2} and {1,3}. So, in a manner of speaking, the set {0,2} is the zero of the quotient group.

3. Apr 30, 2012

### Diffy

Did you mean: "since 0 +1 = 1 and 2 + 3 = 1 under mod 4 addition"?

Yes I absolutely meant that. Thanks.

I guess my problem is seeing where this operation comes from.

The wiki says this:
Does this mean that because 0 & 2 "represent" {0,2} and 1 & 3 "represent" {1,3} when we do {0,2} + {1,3} we can substitute the representatives? So {0,2} + {1,3} becomes 0+1 and since 0+1 = 1 and 1 represents {1,3} the answer is thus {1,3}? Likewise if we chose other representatives 2 + 3 = 5 mod 4 = 1 we still get the representative of {1,3}.

Thanks for taking the time to explain this.

4. Apr 30, 2012

### SteveL27

Note that when you define coset addition, you have to PROVE that it's independent of the representative you happen to pick. In the textbooks they call this showing that the addition is "well-defined."

5. Apr 30, 2012

### Stephen Tashi

That is basically correct.

What represents what is somewhat complicated in this example since so many different groups are involved.

If the group we are talking about is "the integers mod 4 under the operation of addition" then it has 4 elements, so you could denote it as the set ${A,B,C,D}$. If you want to make clear how this group is derrived as a quotient group of the integers, you have to say what subset of the integers corresponds to $A$ etc. When people denote the integers mod 4 by the 'residue classes' { 0, 1, 2 ,3 }, those individual symbols can be regarded as mere symbols or they can each be regarded as a set of numbers (e.g. '2' = all integers of the form 4 n + 2 ) or they can be regarded, in a hybrid fashion, as "representatives". The "representative" denotes a set of numbers, but the symbol for the representative also denotes a particular element of a group that is the "parent" group of the quotient group.

The group defined by " (the integers mod 4) mod 2" has two elements, so you could denote them as the set {A,B}. Or you could denote them by representatives. If you decide to denote the elements by representatives, the question becomes "how did I denote the elements of the parent group?". If you denoted the elements of the integers mod 4 as {0,1,2,3} then the representatives for the two elements quotient group could be chosen as {0,1} since 0 and 1 are elements of the parent group. I suppose you could pick a different pair representatives such as {2,3}. However, we must recall that '1' in the "the integers mod 4 under addition" is technically not the same as the integer 1 in the group of integers under addition. In "the integers mod 4", 1 is a representative for a infinite set of integers. So in "(the integers mod 4) mod 2", the '1' is a representative of a set of two things {1,3} and each of those things is an infinite set of integers.

The use of representatives is, or verges on being, an abuse of notation. It's using a element of one group to denote a subset of elements in that group.

That example in the Wikipedia article on quotient groups is using a mixed notation. It uses a symbol like '1' as a representative when it employs it in the integers mod 4. But when it speaks of subsets of the integers mod 4, it writes out the entire set instead of using one representative for it.

Incidentally, quotient groups are one of the reasons that normal subgroups are important in group theory. If you partition a group G into, say, 3 arbitrary sets {A,B,C} with "representatives" {a,b,c} then you will, in general, find that you haven't defined any sort of group. Trying to do group arithmetic with the representatives won't produce consistent answers and the product of the sets A and B might not even be a set in {A,B,C}. SteveL27 has remarked on this.

Normal subgroups are important since their cosets define partitions of the group that work for defining quotient groups and using representatives.

6. Apr 30, 2012

### Bacle2

For yet another perspective, if you have a quotient A/B , where * is the operation, then a,a' in A are identified iff a*a'-1 is in B (by def. of cosets). In our case, A is Abelian, with *:=+, standard addition.

Then 4Z/{0,2} identifies :1~3, and 0~2 , since 3-1=2 is in {0,2}, same for 2-0. Notice
that 1~1, 3~3,2~2, 0~0. But now we have addition Mod{0,2}

Since these are equivalent, as was said, you can choose a rep. Your choices are:

{[1],[2]}, {[1],[0]}, {[3],[0]},{[3],[2]}

7. Aug 14, 2012

### Diffy

Sorry to resurrect an old thread. But I did originally start it and I always struggle with this stuff.

The context of this question came from trying to understand what a Vitali set is.

I have been looking at the quotient group R/Q. I have been trying to wrap my head around this particular group.

I get that Q is basically the identity element of R/Q

Now for every irrational number in R is there a distinct coset that is formed?

But the other cosets, or members of R/Q, what are they?

I first thought that for each irrational number x in R the coset x + Q is unique.

But then I thought about the number pi. And wouldn't pi + Q = 2pi + Q? Or for any two number q1 and q2 in Q wouldn't you have that pi*q1 + Q = pi*q2 + Q?

Can someone shed some light on R/Q and help me better understand it? Is there another group that this is isomorphic to that I would possibly be familiar with?

Thanks,

-Diffy

8. Aug 14, 2012

### Bacle2

If you consider the Abelian groups, (R,+) and (Q,+) , the quotient group

(R/Q ,+)is defined to be the collection of classes r+Q , where we define

(r+Q)+(r'+Q):= (r+r')+Q . Two classes (r+Q) and (r'+Q) are equivalent iff:

r-r' is in Q . From this we have that the classes are :

Q , and (a+Q) , where a is irrational. This is because to have two irrationals have

a rational difference, they must be in the same class, i.e., a_1+Q ~a_2+Q, then

a1-a2 is in Q, or a1=a2+Q.

I can't think offhand of a group isomorphic to R/Q which is manageable.

Notice that if pi+Q ~ 2pi+Q , then we need 2pi-pi =pi to be in Q

where two classes r+Q a