Cosets of group

when its said "the subset of all left coset of some subgroup" is it meant that i spam every g from G to subgroup N? What is it Im getting then and how come they make disjoint sets that togather complete group G?

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>>8263996
I also struggled with the intuition behind them when I first took this course. I'll share what other anons explained to me now, and then provide an explanation of how I think about them when I get home and have the time

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>>8264117
Kek and then there's this one

>>8264117
This definition via equivalence classes also doesn't click to me. I was first introduced to fibers of homomorphisms, that made sense. I also don't have problem with generally understanding quotient group (I guess it's because of topology reference)

You might want to try to reword your questions to make it a bit clearer what exactly it is you're asking.

>Representative of coset gN is any element of that coset? Is that true?
Any element g' in the left coset gN is a representative of that coset and can be used to generate the coset by leftmultiplying it with all elements of the subgroup N.

>I understand that if N is kernel of some homomorphism, then gn = g1N for any g1 in G.
I'm not sure what you're trying to say here. If g and n are elements, then gn is an element, but g_1N denotes a coset.

>>8264129
The fibers are the cosets, and are elements of G/N

>>8264148
Shouldn't you say that element of left coset, gN, is written in form of gn, n from N. You say element g' from coset and that is what i don't understand. Does it mean then that each left coset is uniqly determinated by element g from G? That has makes sense. And as you say, he generates coset gN.

About the second remark:
>If g and n are elements, then gn is an element, but g_1N denotes a coset.
Can you explain this? Did you mean g_1n is element of coset and g_1N denotes a coset?

>>8264162
Of course, if N is kernel. I suppose that I can move away from kernel in quotient group via these normal groups? (I still didn't reach them, i want to understand this concept first)

>>8264169
All that is required of N is that it is a normal subgroup. But yes that also means N is the kernel (since, if 1 is the identity in G, 1*N =N). I guess it's two sides of the sane coin, really

I think the thing that confuses me most is theorem:
"Set of all left cosets of N in G forms a partition of G". How is it certian that g1N and g2N won't have non trivial intersection? I have seen proof but its one of those "double inclusion proofs"

>>8264192
It's because we have defined an equivalence relation, and if you recall equivalence relations partition the set/group into disjoint subsets (the cosets) which are called equivalence classes.

Note it isn't necessary we have a normal subgroup; any subgroup of G will do. The nice thing about normal subgroups is that gN=Ng for all g in G

>>8264201
I understand how fibers make disjoint set that cover group G. But when I leave realm of homomorphism and keep in mind only arbitary subgroup of G i don't understand how there's no interceptions. In Dummit's Abstract algebra, cosets are not introduced as equ.classes, so I guess that's whats wrong with me

>>8264093
The best way to understand anything is to consider examples
The plane R^2 is a group for component-wise addition, the line L = {(x,0), x in R} is a subgroup. Its cosets are the translates (u,v) + L = {(x,v)|x in R} for v in R

>>8264217
Well just write down what it means: Assume g1N and g2N have nonempty intersection. Then there are n1,n2 in N s.t g1n1 = g2n2. Then g1/g2 = n2/n1, hence g1 = g2 * (n2/n1), hence g1 is in g2N and conversely, g2 is in g1N, hence g1N = g2N.

>>8264192
For simplicity, I'll assume the subgroup is normal, so [math]gN=Ng[/math]

Firstly, note that a representative g can be used to construct the coset gN by left-composing g with all elements of N. [math]gN = \{g\circ n\mid n\in N\}[/math]
It's also important to remember that a coset gN is a set of elements, not a subgroup.

If g ∈ N, then since N is a subgroup it must be closed under composition, and since all elements in gN are of the form g⋅n, they must must also be in N. Since e ∈ N (identity), g⋅e = g is in gN. Furthermore, two distinct elements n' and n'' of N composed with g cannot be equal, since g⋅n' = g⋅n'' would imply n'=n''. Thus the set gN covers the entire subgroup N.

If g is outside of N, then gN will be completely distinct from the elements in N. Assume to the contrary that there is an element n such that g⋅n ∈ N. Then [math]g⋅n⋅n^{-1}=g[/math] ∈ N -> contradiction.
Since g is composed with |N| distinct elements in N, gN must contain exactly |N| elements for the same reason as in the previous example. Thus now we have shown that any coset generated by an element outside the subgroup is disjoint from N and has the same size as N.

Let h ∈ G be another element not in N and not in gN. Then the coset hN must be disjoint not only from N, but also gN. Assume to the contrary that there is some n ∈ N such that h⋅n = g⋅n', where g⋅n' is an element in gN. But then [math]h = g⋅n'⋅n^{-1}[/math], which would imply that h had to be an element of gN to begin with by definition of gN (N closed under composition and inverses). Thus if h is not in gN, then the coset hN must be disjoint from gN.

This shows why the cosets are disjoint and of the same size as N, and thus partition G into subsets of size |N|.

>>8264229
Thank you, I understood it perfectly now :). Proof is really nice but my intuition was wrong

just for revision:

It is quite clear that if I look at g1 from gN, since N is a subgroup, g1n1 will belong in gN (for any n1 in N. Because g1 = gn, and if i multiplay it with some n2 i get g1n2 = g(nn2) which belongs to gN).

So we have all these g1,g2,g3... from gN set and they all have in common some (not same) element of N that gives us i.e. g1 = gn. I think that equivalence relation mainly comes from N being closed under its operation. It's easy to check that gN "sticks" some other g-s inside it.

Now, in order that all these gN (which we can now look as quotient G/~ where ~ is former eq. relation) become group, we must declare operation on them. We do it by: for u,v in G, uN*vN = (uv)N. Problem is that this operation is well defined if certian property stands, and that is: gng-1 belongs to N for all g and n. Is this property directly connected to some property of subgroup N or generally to group G?

>>8266077
>Problem is that this operation is well defined if certian property stands, and that is: gng-1 belongs to N for all g and n. Is this property directly connected to some property of subgroup N or generally to group G?
What you're talking about is a normal subgroup, which is defined as a subgroup that is invariant under conjugation, i.e.
[math]gng^{-1}\in N[/math]
and as a consequence [math]gN=Ng[/math]
The subgroup N being normal is indeed a requirement for it being able to form the kernel of a homomorphism.

>>8266092
I found it a bit exhausting to keep track of stabilizers, normalizers, centralizers. But if I remember correctly, normalizer will have something to do with this normal subgroup, right?

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>>8266092
>The subgroup N being normal is indeed a requirement for it being able to form the kernel of a homomorphism.
and importantly, kernels are always normal

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>>8264117
very surprised to see you saved my short explanation from over a month ago, did you find this useful?