Attempting to better understand the group isomorphism theorm

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    Group Isomorphism
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Discussion Overview

The discussion centers on understanding the group isomorphism theorem, particularly the relationship between a homomorphism and the induced isomorphism between the quotient group G/Ker(p) and the group H when the homomorphism p is onto. Participants explore the implications of this theorem and seek clarification on specific aspects of the mapping and the properties of cosets.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asserts that if p(a) = p(b) for elements a and b in G, then aK = bK in the factor group, seeking an intuitive understanding of this claim.
  • Another participant explains that since p is a homomorphism, p(ab^{-1}) is the identity in H, which implies that ab^{-1} is in the kernel of p.
  • A different participant elaborates that aK is equal to p^(-1)(p(a)), arguing that if p(a) = p(b), then aK = bK follows from this relationship.
  • One participant questions the assertion that aK equals p^(-1)(p(a)), suggesting that applying the mapping and then the inverse mapping would yield a = p^(-1)(p(a)).
  • Another participant clarifies that there is no inverse mapping in this context, as the mapping is only surjective, and discusses the concept of inverse images in relation to group homomorphisms.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the mapping and the concept of inverse images. There is no consensus on the interpretation of the relationship between cosets and the mapping induced by the homomorphism.

Contextual Notes

Some participants highlight the need for a clearer understanding of basic set theory and functions, particularly regarding the concept of inverse images versus inverse mappings.

PsychonautQQ
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The homomorphism p:G-->H induces an isomorphism between G/Ker(p) and H (if p is onto). I am trying to understand why this must be true. I understand why these groups have the same magnitude and so a bijection is possible, but there is something that I am not able to understand.

What seems to be true but I don't understand why is that if a and b are elements in G, that if p(a) = p(b) then aK = bK in the factor group. Is this true? It seems like it should be. If so, then can somebody help me understand why it is true on an intuitive level?

Sorry, I realize that this is a broad and poorly phrased question, hoping somebody can see through my lack of competent communication and give me some insight here.
 
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Since ##p ## is a homomorphism ##p(ab^{-1}) ## is the identity in H so ##p(ab^{-1}) ## is in the kernel of ## p ##.

So ##ab^{-1} = k## for some ##k ## in K
 
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put another way, aK is exactly equal to p^(-1)(p(a)), since (in combination with lavinia's answer): for x in K, p(ax) = p(a)p(x) = p(a)e = p(a);

so if p(a) = p(b) = y, then aK = p^(-1)(p(a)) = p^(-1)(y) = p^(-1)(p(b)) = bK.
 
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how does aK = p(-1)^(p(a))? wouldn't p^(-1)(p(a)) = a? applying the mapping and then the inverse mapping?
 
there is no inverse mapping. only bijective mappings have inverse mappings, and yours is only surjective. in general, the notation f^(-1), denotes not the inverse mapping from points of the target to points of the source, but rather the inverse image mapping from subsets of the target to subsets of the source.

I.e. in general, as here, if f:S-->T, is a mapping, then f^(-1) of a subset Y in T, means the subset X of S consisting of all points that map into Y.For a one point subset say y in T, f^(-1)(y) means the subset of all points of S that map to y. For a group homomorphism f:G-->H, the kernel is the subset f^(-1)(e), i.e. the subset of all points of G that map to e, and if y is a point of H, and x is any point of G with f(x) = y, then the coset xK equals the subset f^(-1)(y) of all points of G that map to y.

This material on inverse images of maps that you seem to be missing is more elementary than group theory. Maybe you need to review basic theory of sets and functions.
 
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