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Why is a homomorphism the way it is?

  1. Sep 30, 2008 #1

    tgt

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    Is the reason why the homomorphism is defined the way it is so that we can be sure that the image is a group?

    Could you get a map which is not a homomorphism but still have the image as a group? This question is equivalent to asking whether in the domain group, elements can multiply differently so that it can still form a group. Can it?
     
    Last edited: Sep 30, 2008
  2. jcsd
  3. Sep 30, 2008 #2
    Below is an algorithm (one of many) to find a mapping whose image is a group but the mapping itself is not homomorphism.

    (Surgective)
    1. Pick two finite group A, B, where the size of A is bigger than B and B is not trivial group.
    2. Find a random "onto" mapping f from A to B such that dom f = A and ran f = B.
    3. Check if it is a homomorphism mapping.
    4. If it is a homomorphism mapping, Go to step 2. Otherwise retrieve the answer.

    If the step 2-4 is not working out for some trials, go to step 1 and repeat the procedure.

    (Injective)
    1. Pick two finite group A, B, where the size of A is smaller than B and A is not trivial group.
    2. Find a random "into" mapping g from A to B such that dom g = A and ran g = subgroup of B.
    3. Check if it is a homomorphism mapping.
    4. If it is a homomorphism mapping, Go to step 2. Otherwise retrieve the answer.

    If the step 2-4 is not working out for some trials, go to step 1 and repeat the procedure.
     
    Last edited: Sep 30, 2008
  4. Sep 30, 2008 #3

    tgt

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    ok. I see. So we can have maps between groups that aren't homomorphisms. How widely used are those maps?

    As an example, take two different finite groups generated by the same elements and with the same order. Claim: There exists a map defined element to element which is not a homomorphism. i.e a->a, b->b etc. But the image, domain and codomain are all groups. It seems these maps are not very useful and that is why they are not used?

    It also seems possible to find a group with more than one way of multiplying elements together.

    A homomorphism is just a way map that preserves multiplication between two groups. And 'multiplication' is the central theme in all four group axioms. So a group is a set with a certain way of multiplication of elements.

    It seems that the more natural thing to do is to first define an isomorphism (a very intuitive concept) and then make the condition of the map weaker and we have a homomorphism.
     
    Last edited: Sep 30, 2008
  5. Oct 1, 2008 #4
    tgt, I think you are missing the point. When you use the word "group" that means that you already know how to multiple elements together, its part of the data for that group. The actual labels that are assigned to the elements is irrelevant: you can all the elements a, b, c or 1,2,3 or whatever you like the whole point is how they multiply together to give another element in the group. And that's exactly what a homomorphism tries to capture: how similar are two groups up to relabeling of the elements. I.e. if if I have two groups one with elements a, b ,c with a given way of multiplying and another group with elements 1, 2 ,3... with a given way of multiplying can I relabel my a, b, c... with 1, 2, 3 in such a way so as to obtain the second group? The map which tells you how to relabel is called a homomorphism.
     
  6. Oct 2, 2008 #5

    tgt

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    What you've said justifies what I wrote earlier which is that 'It seems that the more natural thing to do is to first define an isomorphism (a very intuitive concept) and then make the condition of the map weaker and we have a homomorphism.' The bijection captures the general idea of equivalence between sets and the homomorphism is an added notion applying only in an algebraic context, highlighting the multiplication of elements in sets. So two group are homomorphic if and only if there exists subgroups in each group which have the same way of multiplying together all elements.
     
  7. Oct 2, 2008 #6
    Why is an isomorphism any more intuitive then a homomorphism?? An isomorphism is just a homomorphism which has an inverse which is also a homomorphism. In fact categorically, that's how an isomorphism is defined, so you can't even define an isomorphism until you have the notion of a homomorphism. I think your mistake is that you think an isomorphism is just a map between sets without regard to the multiplicative structure - that is false.

    I think confusion will stop once you actually write out the definition of an isomorphism between two groups. You will see you won't be able to do it without defining a homomorphism first.
     
  8. Oct 3, 2008 #7

    tgt

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    Then you have clearly misunderstood me. Have a read of the 3rd post.

    I think it is much easier to imagine a pair of isomorphic groups then a pair of groups which are only homomorphic (and not isomorphic). Generalisations usually come later (in a person's thought process) and homomorphism is a generalisation of isomorphism.
     
  9. Oct 3, 2008 #8

    morphism

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    If this feels more natural to you, then by all means, think of it this way. There's really nothing wrong with what you're saying.
     
  10. Oct 4, 2008 #9

    tgt

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    This thread came up because of a memory of question I asked a prof.

    Me: "Does a map between two groups must be a homomorphism"
    Prof: "Yes"

    But there are maps between two groups that aren't homomorphisms as shown in this thread so the Prof should have said instead "For 'practical' purposes, yes" .
     
  11. Oct 4, 2008 #10

    morphism

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    I find it very hard to believe that a professor would give that answer!
     
  12. Oct 4, 2008 #11

    Hurkyl

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    Those aren't maps between groups. Those are maps between their underlying sets. A map of sets need only be a function. A map of groups must be a homomorphism. (By definition of the word 'map'!)
     
  13. Oct 4, 2008 #12

    mathwonk

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    homomorphisms are defined to be able to compare groups. there is a partial order on the family if all groups, in the sense that G > H iff there is a homomorphism from G onto H.
     
  14. Oct 5, 2008 #13

    tgt

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    Can the same idea be applied to sets in general? Hence functions are defined to compare sets?
     
  15. Oct 6, 2008 #14

    tgt

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    If what you say is true then why do we need terminology such as homomorphism? We can instead say a map between two groups and people should automatically know what the map has to satisfy.

    A map which is a function is just a relation between two points. The underlying sets can be anything if they satisfy the axioms for a group then the map is between two groups. If the two sets also satisfy the axioms for a vector space then that map can also be called a map between two vector spaces.
     
  16. Oct 6, 2008 #15
    Interesting this is how Artin does it in his Algebra Book.

    He first introduces isomorphism, then moves on to homomorphisms.
     
  17. Oct 7, 2008 #16

    tgt

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    That how Hungerford does it as well.
     
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