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Homework Help: Show Quotient Groups are isomorphic

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that Z18/M isomorphic to Z6 where m is the cyclic subgroup <6>
    operation is addition

    3. The attempt at a solution
    M = <6> , so M = {6, 12, 0}

    I figured I could show that Z18/M has 6 distinct right cosets if I wanted to do M + 0 = {6, 12, 0} M + 1 = {7, 13, 1} M + 2 = {8, 14, 2} etc and after M + 5 , they would start to repeat

    After that I do not know how you show isomorphism. I know it has to satisfy three conditions, injective, surjective, and homomorphism, but I do not know what the function is even, so I'm assuming it is f(a) = M + a for a an element of Z18

    so injective : if f(a) = f(b) so ( M +a = M + b ), then a = b.
    M + a = {6 +a, 12 + a, a} and M + b = {6 + b, 12 + b, b}
    then 6 + a = 6 + b
    12 + a = 12 + b
    a = b , so I guess I need to show that each of those three equations show a = b, and I will use cancellation of 6, and 12 and we get a = b, so f is injective (but I don't understand how it is injective

    surjective: let a be arbitrary element of Z18, then f(a) = M + a = {6 + a, 13 + a, a} so a would be in Z18/M , so it is surjective

    homomorphism: f(a + b) = f(a) + f(b) , so
    M + ( a + b) = M + a + M + b = f(a) + f(b) this result came from definition of a quotient group from my book which said that G/N (N a normal subgroup) is a group under the operation defined by(Na)(Nc) = N(ac)

    Now my book gave a hint saying show that 1 +M generates all of Z18/M, and so Z18/M is cyclic and then show that the order of (1 + M) is 6, however, I do not quite understand how you would show that a set has order since 1 + M = {7, 13, 1} right?
    much less showing that 1 + M generates all of Z18/M. thanks
  2. jcsd
  3. May 5, 2012 #2
    There are three groups that you're looking at here; [itex]Z_{18}[/itex], [itex]Z_6[/itex], and the quotient group [itex]Z_{18}/M[/itex].

    Which of these two groups are you meant to show are isomorphic? In your work, what is the "domain" of the isomorphism that you're trying to construct?

    If I were you, I would focus on understanding the quotient group a little better before you try to solve the problem. What are the elements of the quotient group? Does an element of the quotient group have multiple names? Can you make a Cayley table for this group? Can you find a homomorphism from [itex]Z_{18}[/itex] onto [itex]Z_{18}/M[/itex]? what is the kernel of this homomorphism?
  4. May 6, 2012 #3
    I'm meant to show the quotient group Z18/M is isomorphic to Z6.

    The elements of Z18/M are
    M + 0 = {0, 6, 12}
    M + 1 = {1, 7, 13}
    M + 2 = {2, 8, 14}
    M + 3 = {3, 9, 14}
    M + 4 = {4, 10, 15}
    M + 5 = {5, 11, 16}

    so, I need a function f that takes z18/m to Z6 , so the domain would be the set of Z18/M which are the elements above.

    The elements of the quotient group can only have one name, since the quotient group is the set of distinct right cosets.

    Yes a Cayley table shows that M + 0 is 0
    M + 1 is 1
    M + 2 is 2
    M + 3 is 3
    M + 4 is 4
    M + 5 is 5
    where 0,1, 2, 3, 5, would be the elements from Z6
    I've attached the Cayley table I made.

    I'm sure we could find a homomorphism from Z18 onto Z18/M
    f : Z18 → Z18/M where f(a) = M + a
    then f(a+b) = M + (a+b) = (M +a) + (M + b) = f(a) + f(b)
    the kernel would be the set of elements in Z18 that get mapped to the identity element of Z18/M which is the coset M
    Then the kernel is {0, 6, 12}

    so I suppose since we need to show Z18/M is isomorphic to Z6 we could use the Cayley tables since it would seem each element of Z18/M corresponds to one element from Z6 and we basically just relabel the table of Z18/M with the elements they correspond to from Z6 and we have the same tables.

    Attached Files:

  5. May 6, 2012 #4
    Ok, good. It looks like you've found the correct mapping [itex]\phi:\mathbb{Z}_{18}/M\rightarrow \mathbb{Z}_6[/itex]. It's bijective (which I think we can agree is obvious). It's also a homomorphism, which may not be as obvious. Can you show that [itex]\phi[/itex] is a homomorphism?

    While I think you've accomplished the goal of the problem, it may not be exactly what the problem was supposed to help you understand. If I had to guess, you've recently covered the First Isomorphism Theorem? See if you can find a surjective homomorphism [itex]\Phi:\mathbb{Z}_{18}\rightarrow \mathbb{Z}_6[/itex] with [itex]Ker(\Phi)=M[/itex]. Try to see how this would give you an alternate way to solve the problem. The various Isomorphism Theroems are a big deal, and it's helpful to see how they work in these "easy" cases so that we believe that they work in the more difficult cases.
  6. May 6, 2012 #5


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    If you have proven that [itex]Z_6[/itex] is the unique abelian group of order 6, then you can also just note that [itex]Z_{18}/\langle 6 \rangle[/itex] is an abelian group of order 6. This is another useful technique in finite group theory.
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