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Math Help - Quotient Groups - Infinite Groups, finite orders

  1. #1
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    Quotient Groups - Infinite Groups, finite orders

    I have a question: when I find a quotient group, I actually find the group of all cosets? I mean, every element in a quotient group is actually a coset?

    For example, I have a confusing question:
    Let (Q,+) be the rational group, (Z,+) be the integer a subgroup of Q.
    Prove that every element in Q/Z has a finite order. (Does this mean I need to prove that every coset has a finite number of elements? or, what?)

    Anyway, I googled this question, and I found a proof for that:

    let z+q be an element in Q/Z, while q \in Q and z \in Z,
    then q=\frac{a}{b}, while a,b \in Z and b!=0.
    Then, b*(z+q)=bz+qb=bz+a \in Z

    therefore, |z+q|=b <<Why is that?! What is the neutral element in Q/Z? I don't understand this>>

    Please help me,

    Thank you very much
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  2. #2
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    Quote Originally Posted by adam63 View Post
    I have a question: when I find a quotient group, I actually find the group of all cosets? I mean, every element in a quotient group is actually a coset?
    Yes.

    For example, I have a confusing question:
    Let (Q,+) be the rational group, (Z,+) be the integer a subgroup of Q.
    Prove that every element in Q/Z has a finite order. (Does this mean I need to prove that every coset has a finite number of elements? or, what?)
    No, an element of a group is said to have "finite order", if you can reach the neutral element by multiplying( i.e. applying the group operation) the element to itself a finite number of times.
    For example the element 1 in Z has NOT finite order, because you can never reach 0 just by adding 1 to itself.
    In fact no element in Z except 0 itself has finite order.

    Anyway, I googled this question, and I found a proof for that:

    let z+q be an element in Q/Z, while q \in Q and z \in Z,
    then q=\frac{a}{b}, while a,b \in Z and b!=0.
    Then, b*(z+q)=bz+qb=bz+a \in Z

    therefore, |z+q|=b <<Why is that?! What is the neutral element in Q/Z? I don't understand this>>
    The neutral element in Q/Z is the coset 0+Z=Z. In general the neutral element of any quotient group G/N is the coset 0+N=N.

    If you have any coset \frac{a}{b}+\mathbb{Z}, b>0 you can reach the neutral element by adding it to itself b-times, because
    b\cdot\frac{a}{b}+\mathbb{Z}=a+\mathbb{Z}=0+\mathb  b{Z}.
    Therefore every element in Q/Z has finite order.
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  3. #3
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    Oh, beautiful, I understand it

    Only one last question:

    if I multiply b*(a/b+Z), shouldn't I get: a+bZ, which is a different coset?

    *Now I understand - it's because you only add Z again and again... which is an infinite group, and you will always have the 'full' Z, no matter how many times you multiply it
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  4. #4
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    Quote Originally Posted by adam63 View Post
    Oh, beautiful, I understand it

    *Now I understand - it's because you only add Z again and again... which is an infinite group, and you will always have the 'full' Z, no matter how many times you multiply it
    Exactly. The notation b*(a/b+Z) can be a bit confusing, so i edited out the paranthesis now. What is meant is indeed just adding the whole coset a/b+Z b-times to itself. You can also operate only on a representative a/b of the coset and then you get a representative b*a/b for the new coset. Note that a coset can have several representatives. Every element of the coset is a representative and two representatives x,y are equivalent, if x-y is in Z.
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