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Normal Subgroups and Isomorphisms

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the subset H = {0,3} of the group Z6.
    (i) Prove that H is a normal subgroup of Z6.
    (ii) Prove that Z6/H is isomorphic to Z3. (Give an explicit isomorphism)

    2. Relevant equations
    In order to be a subgroup H must be closed under +6, have an identity, and have an inverse for each element.
    In order to be a normal subgroup, the left cosets must equal the right cosets of H in Z6.

    3. The attempt at a solution
    (i) H is closed under +6
    I wrote out the group table:
    0 +6 0 = 0
    3 +6 0 = 3
    0 +6 3 = 3
    3 +6 3 = 0
    Therefore, H is closed.

    the identity, e, is also in H
    e = 0
    0 +6 0 = 0
    3 +6 0 = 3
    Therefore, e is in H.

    the inverse of a, a-1, is in H
    0 +6 0 = 0
    3 +6 3 = 0
    Thus, a = a-1
    Therefore, a-1 is in H.

    Since all three conditions hold, H must be a subgroup of Z6.

    Is H a normal subgroup of Z6?
    left cosets:
    aH =
    0 +6 H = {0,3}
    1 +6 H = {1,4}
    2 +6 H = {2,5)

    right cosets:
    Ha =
    H +6 0 = {0,3}
    H +6 1 = {1,4}
    H +6 2 = {2,5}

    The left cosets = the right cosets of H in Z6.
    Thus, H is a normal subgroup of Z6.

    Is the above work correct?

    Part (ii) I do not know where to even begin.
    What does Z6/H mean? Is it possible to draw a group table for it?
    I would really appreciate some heavy guidance on this portion.

    Thank you so much for any help you can provide me!
     
  2. jcsd
  3. Apr 25, 2012 #2
    If a group is Abelian then all its subgroups are normal, because $$gxg^-1 = x$$ always. That saves some work.

    If you're not sure what Z6/H means, you should review that in your book or class notes.

    It's the set of cosets, endowed with a particular group operation that will be described in extreme detail in any algebra text. It's a really important construction, so it's worth the time to understand it.
     
  4. Apr 25, 2012 #3
    Thank you for your response! I suspected all Abelian groups would share that property. As for part (ii) I have looked it up extensively and read much on the topic, however, I am still struggling to fully understand it. Perhaps I just need to see it worded differently.

    Is it simply ={{0,3},{1,4},{2,5}}?
     
  5. Apr 25, 2012 #4
    Yes those are the elements of H. But what is the group operation? Well the trick is that we define addition of cosets by taking a reprepresentative from each coset. So to add {1,4} + {2,5} we can just add 1 plus 2 and the answer is whatever coset 1 + 2 is in, where '+' here means the original group operation of addition mod 6. (Which coset is that?)

    But what if I picked a different representative? Would I get the same coset? You have to prove that the definition of + for cosets is "well defined," in the sense that it's independent of the particular choice of representatives.

    Are you working from a book? This is explained in copious detail in every book on abstract algebra.

    By the way, what is Z6? You should not be doing this problem until you understand the construction of Z6. It's the first and simplest example of the factor group construction. It's the integers "modded out" by the multiples of 6.

    If you understand that example, then you'd know what you need to know for this problem. But if you don't understand the factor group construction of Z6, it's essential (or at least extremely helpful) to understand that first.
     
    Last edited: Apr 25, 2012
  6. Apr 26, 2012 #5
    Thank you again, your input has been extremely helpful.
    I am working out of a text right now, as part of my undergrad course in abstract algebra. However, the course I am in now has only covered the very beginning of the material. Factor groups have not been covered and sadly appear much later in the text, where it is surrounded by terms and definitions I do not yet understand. That's why google and helpful posters like you are much appreciated :).

    Let me see if I fully understand what you have written:
    I note that when adding (mod 6) any two elements of Z6/H = {{0,3},{1,4},{2,5}} in the traditional way I get a double digit answer: {0,0}, {1,1}, {2,2}...which makes sense. So, instead I need only choose a "representative" from each coset. Therefore the result is not {0,0} but {0,3}, {1,4}..etc.

    So, if I were to write out the group table I would have:
    1st row: (0,3) (1,4) (2,5)
    2nd row: (1,4) (2,5) (0,3)
    3rd row: (2,5) (0,3) (1,4)

    Therefore, the isomorphism to Z3 becomes much more clear where
    (0,3) ---> 0
    (1,4) ---> 1
    (2,5) ---> 2

    Or have I completely misunderstood you haha
     
  7. Apr 26, 2012 #6
    It is my understanding that Z6 is the group formed by {0,1,2,3,4,5} under addition mod 6. For example, 3 + 5 = 2.
     
  8. Apr 26, 2012 #7
    That's sufficient for some purposes, but not for abstract algebra class. We have to "re-learn" the definition of Z6 when we first study group theory, by understanding Z6 as an example of a much more general construction of factoring a group by a normal subgroup.

    If you take the integers Z as a group, consider the subgroup H = {multiples of 6}.

    Now Z6 = Z/H. It is essential that you thoroughly understand this example before proceeding any further.

    Have you seen what's commonly called the First Isomorphism Theorem?

    How is it possible for you to have been assigned the question you posted without knowing the basic factor group construction? It's a question about factor groups so you need to know what they are before attacking this problem.

    I would like to be helpful but I feel inadequate to the task of logically presenting material that's covered in every book on abstract algebra. But we should start with Z6 = Z/6Z where 6Z denotes the multiples of 6. That's the first thing to understand here.

    Can you at least tell me what book you're working from? I'm just curious.
     
    Last edited: Apr 26, 2012
  9. Apr 26, 2012 #8
    You've got the right idea, but it would be a lot better if you understood that Z6 = Z/6Z to start with. But that's the right isomorphism.
     
  10. Apr 26, 2012 #9
    I would like to ask the same question! haha. I think my professor may have added the problem in as a challenge. The text I have is called A First Course in Abstract Algebra (7th ed.) written by John Fraleigh. It's a well written text and I have, for the most part, found it very helpful.

    I understand factor groups a lot better now. It seems to be a way of defining groups in relation to others, rather than in isolation.
    So Z6 = Z/6H =
    {{...-6,0,6,12...}
    {...-5,1,7,13...}
    {...-4,2,8,14...}
    (...-3,3,9,15...}
    {...-2,4,10,16...}
    {...-1,5,11,17...}}
    ?
     
  11. Apr 26, 2012 #10
    Yes, exactly. We take the subgroup 6Z = {multiples of 6} and consider its six cosets. We make the cosets into a group by saying that the addition of two cosets is defined as the coset containing the sum of any two members of each respective coset. Then we prove that this coset addition is well-defined, by showing that it doesn't matter which representative we choose from each coset.

    Now we can define Z6 = {[0], [1], [2], [3], [4], [5]}

    where the notation [n] stands for the coset that n is in.

    So for example [3] = [9] = [33].

    Now we can say that [2] + [3] = [5], once we have shown that we'd get the same answer if we did [8] + [33] = [5], having already shown that coset addition is independent of which representative we choose.

    Now in casual conversation we say that Z6 is {1, 2, 3, 4, 5} with "addition mod 6" but what we really mean is that we have in mind the coset construction, and instead of writing [4] we agree to write 4 -- as long as we remember that the element of Z6 that we call 4 is really [4], the coset of 6Z containing 4.

    In other words the elements of Z6 are not integers, but are rather cosets of the normal subgroup 6Z. (As we showed earlier, subgroups of Abelian groups are always normal.)

    Now the next bit is that we can define a homomorphism from Z to Z6 by mapping each n to [n]. And what is the kernel of this homomorphism? It's ... drum roll ... 6Z, or the subgroup of Z consisting of the multiples of 6.
     
    Last edited: Apr 26, 2012
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