I Understanding Quotient Groups in Abstract Algebra

[Wledig]

So I'm just beginning to study abstract algebra and I'm not sure I grasp the definition of a quotient group, I believe it probably has to do with the book providing little to no examples. In trying to come up with my own examples, I imagined the following:
Consider the Klein four group, if we take the subgroup (e,a) and apply the defintion we ought to get

e(e,a) = (e,a)
a(e,a) = (a,e)
b(e,a) = (b,c)
c(e,a) = (c,b)

So K4/(e,a) = {(e,a), (b,c)}. Is that correct? Does the order the elements appear matter?

 

[andrewkirk]

Taking a quotient over a subgroup is essentially removing the structure that the parent group shares with that subgroup. So we are left only with the structure that is not reflected in the subgroup.

The easiest group to use to grasp the notion of a quotient is the set of integers modulo p, as a group under the operation of addition. That group, written $\mathbb Z_p$ can be written as $\mathbb Z/(p\mathbb Z)$, that is, the quotient of the additive group of integers over the additive group of integers that are multiples of $p$. If you play around with that for a while it should help develop an intuitive understanding of quotients.

Another nice example is the additive group $\mathbb R^2$. If we take the quotient over the additive group $\mathbb R\times \{0\}$ (think of that as the X axis) we get a group that is isomorphic to the Y axis: $\{0\}\times\mathbb R$, which is isomorphic to $\mathbb R$. And taking the quotient over the Y axis essentially gives us the X axis.

In the additive group $\mathbb R^2$, taking the quotient over any line (plane) through the origin gives us a plane (line).

 

[Stephen Tashi]

So K4/(e,a) = {(e,a), (b,c)}. Is that correct? Does the order the elements appear matter?

Those are the correct sets. The order of listing the elements doesn't matter. To polish off your example, you should state the group operation for the quotient group. For example, letting A = {e,a}, B = {b,c} what is multiplication table for the two element group {A,B}. How do we compute results such as A*B or A*A = A^2 ?

The easiest group to use to grasp the notion of a quotient is the set of integers modulo p, as a group under the operation of addition. That group, written $\mathbb Z_p$ can be written as $\mathbb Z/(p\mathbb Z)$, that is, the quotient of the additive group of integers over the additive group of integers that are multiples of $p$.

Let's expand on that example. The elementary way to present "addition mod 7" is to say its is an algorithm performed on the integers involving two steps: 1) Add two integers the usual way 2) Take the remainder of the answer when divided by 7. That viewpoint doesn't involve thinking about a new number system called "$\mathbb{Z}/(7\mathbb{Z})$".

Let's think about "addition mod 7" from the viewpoint of quotient groups. From that viewpoint, the notation "6 + 4 = 3 (mod 7)" is not a statement about an algorithm performed on two integers. To emphasize that fact, it's clearer to use notation like $[6] + [4] = [3]$ so "$[6]$" is not confused with the single integer $6$.

The set [0] = {...-14,-7,0,7,14,21...} is a subgroup of $\mathbb{Z}$ under addition. The coset [6] is formed by adding the single integer 6 to each element of [0]. So [6] = { ...-8,-1,6,13,20,27,...}.

(Incidentally, thinking about modulo 7 arithmetic as a two step algorithm requires knowing how to apply step 2 of the algorithm to negative integers. So a person must remember that -8 = (7)(-2) + 6 and we use the positive remainder 6. If you think about modulo 7 arithmetic in terms of quotient groups, it easier to see that -8 belongs in the set [6].)

The notation [6] + [4] = [3] can be visualized as a statement about sets:
{..-8,-1,6,13,20,27...} + {...,-10, -3 , 4, 11, 18,...} = { ..., -11, -4, 3, 10, 17,...}
However we must define what the "+" means. (It doesn't refer to the union of sets.)

One way to define [6] + [4] is to say it is the set consisting of those integers that result from adding some integer in [6] with some integer in [4].

A different way to define [6] + [4] is to say it is the set formed by adding 6+4=10 to each member of the set [0].

@Wledig, which definition is consistent with the definition of quotient group in your course materials?

The two definitions define the same set, but when we take one as the definition then proving the other describes the same set is a theorem.The definition of a quotient group requires using a normal subgroup, so it would useful to do an example where you take cosets of subgroup that is not normal and see where things get fouled up. That would make it clear why normal subgroups are so emphasized in group theory. For any subgroup H of a group G , you can define the left cosets of H. The problem comes when you try to define a group operation on these cosets in the same manner as you do with a normal subgroup. (For example, if there are 10 cosets of H, you could define a group operation on those 10 sets given complete freedom of how to define such an operation, but you can't necessarily define the group operation in the same manner as specified by the definition of a quotient group.)

 

[fresh_42]

So I'm just beginning to study abstract algebra and I'm not sure I grasp the definition of a quotient group, I believe it probably has to do with the book providing little to no examples. In trying to come up with my own examples, I imagined the following:
Consider the Klein four group, if we take the subgroup (e,a) and apply the defintion we ought to get

e(e,a) = (e,a)
a(e,a) = (a,e)
b(e,a) = (b,c)
c(e,a) = (c,b)

So K4/(e,a) = {(e,a), (b,c)}. Is that correct? Does the order the elements appear matter?

My Eureka moment was, as I've tried to explain the concept. Given a group $G$ and a subgroup $U$, then $G/U=\{\,gU\,|\,g\in G\,\}$ is a set of sets $gU$ which are all equally large and which define an equivalence relation on the elements of $G$ by $g \sim h \Longleftrightarrow gh^{-1} \in U$. So first it has to be shown that this is indeed an equivalence relation (reflexive, transitive, symmetric) and that $|gU|=|hU|=|U|$.

Now we are at the exciting question: Does $G/U$ has a group structure again? So we want to define $gU \cdot hU = ghU$. But this can only be done in a well-defined manner, if $U$ is a normal subgroup. Well-definition here means, because we have sets as elements, namely $gU$, they can be represented by many other elements besides $g$: $gU=(g\cdot u)U$, and we have to make sure, that all these lead to the same result: $g(u)U\cdot (hu')U=ghU$ no matter which $u,u' \in U$ we choose. And therefore we need that $U$ is a normal subgroup. You can't see this immediately, so I suggest to try and find out.

 

[Wledig]

Thanks everyone for replying to the thread, \mathbb{Z}/p\mathbb{Z} really clarified things to me.

And taking the quotient over the Y axis essentially gives us the X axis.

In the additive group $\mathbb R^2$, taking the quotient over any line (plane) through the origin gives us a plane (line).

Not sure I see that, if I understood the definition of the additive group, that would be the group comprising pairs of real numbers under addition, no?

So in doing R2/Y, we would get in general:

(0,y) + (a,b) = (a,b+y)

Where a, b and y can vary.

So you would get a plane, the entire R2 again? What am I doing wrong here? Maybe I'm taking a very operational approach, but it's the only way I can be comfortable with these concepts.

For example, letting A = {e,a}, B = {b,c} what is multiplication table for the two element group {A,B}. How do we compute results such as A*B or A*A = A^2 ?

I believe I should treat it as a cross product?

So:

(e,a)(b,c) = (eb,ac) = (b,b) ?

Either that, or:

(e,a)(b,c) = ( e(b,c), a(b,c) ) = ( (b,c),(c,b) ) ?

Given a group $G$ and a subgroup $U$, then $G/U=\{\,gU\,|\,g\in G\,\}$ is a set of sets $gU$ which are all equally large and which define an equivalence relation on the elements of $G$ by $g \sim h \Longleftrightarrow gh^{-1} \in U$.

I find your definition a bit harder to grasp than the one I'm following: "The quotient group is formed by the set of all left cosets of N (a normal subgroup of G) in G". But certainly easier than the first one I encoutered: "Let \varphi : G \rightarrow H be a homomorphism with kernel K. The quotient group or factor group, G/K (read G modulo K or simply G mod K), is the group whose elements are the fibers of \varphi with group operations defined above: namely if X is the fiber above a and Y is the fiber above b then the product of X with Y is defined to be the fiber product ab."

 

[Stephen Tashi]

I believe I should treat it as a cross product?

No. Don't confuse a set of things with a coordinate system. The order of listing things in a coordinate system matters. Not so in a set.

A = {e,a} and B = {b,c} so AB = { eb, ec, ab, ac} = {b,c,c,b} = {b,c} = B

Or by the other definition of the operation for a quotient group:
A = e{e,a}, B = b{e,a} so AB = (eb){e,a} = (b){e,a} = {be,ba} = {b,c} = B

 

[Wledig]

I see, thanks for pointing that out.

 

[andrewkirk]

So in doing R2/Y, we would get in general:

(0,y) + (a,b) = (a,b+y)

Where a, b and y can vary.

So you would get a plane, the entire R2 again? What am I doing wrong here?

We need to write things out formally to avoid taking wrong steps. Your addition above is of points, but needs to be of cosets. The quotient group $\mathbb R^2/Y$ where $Y\triangleq (\{0\}\times \mathbb R)$ is the set:
$$\{(x,y)+Y\ :\ x,y\in\mathbb R\}$$
We note that $(a,b)+Y=(a,b+y)+Y$ because $(a,b) - (a,b+y) = (0,y) \in Y$, so that
$((a,b)+Y)-((a,b+y)+Y)=Y$ and $Y$ is the zero element of the quotient group. So any two points that have the same x coordinate are in the same element of the quotient (the same coset of Y). Each coset $(x,y)+Y$ is a vertical line through the point $(x,0)$.

If we take a quotient over $X\triangleq \mathbb R\times\{0\}$ then each coset $(x,y)+X$ is a horizontal line through $(0,y)$.

 

[Wledig]

It makes sense now, thanks for the insight.