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Math Help - Groups - Quotient groups

  1. #1
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    Groups - Quotient groups

    Let G be a group, A be a group of all elements in G that have a finite order.

    Prove that the quotient group G/A contains no elements of finite order.

    Hmmm... If I assume that there is a gA in G/A for which there exists a natural n so that |gA|=n , then I get that : gA+...gA=gA (n times)

    How do I continue ? Am I doing this right?

    Thanks!
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    Let G be a group, A be a group of all elements in G that have a finite order.

    Prove that the quotient group G/A contains no elements of finite order.

    Hmmm... If I assume that there is a gA in G/A for which there exists a natural n so that |gA|=n , then I get that : gA+...gA=gA (n times)

    How do I continue ? Am I doing this right?

    Thanks!
    If n(gA)=0A (i.e. gA has order n) then this means that ng \in A (why?). The contradiction now follows.

    (To write it in non-abelian notation, 'If (gH)^n=1H (i.e. gA has order n) then this means that g^n \in H')
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    I see... although I don't really understand why ng is in A - is it because : (ng)A=(e)A=A, or is it because A is the neutral in G/A

    It's a little complicated... Literally, Abstract Algebra...
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    I see... although I don't really understand why ng is in A - is it because : (ng)A=(e)A=A, or is it because A is the neutral in G/A

    It's a little complicated... Literally, Abstract Algebra...
    The identity element of G/A is eA or 0+A, a coset. This coset corresponds precisely to the subgroup A (the elements are the same).

    Now, if g+A has finite order then there exists an n such that n(g+A) = ng+A = 0+A. Thus, (ng)-0 \in A \Rightarrow ng \in A.

    (The bit where you go ` (ng)-0 \in A' holds because a+A = b+A \Leftrightarrow a-b \in A)

    In multiplicative notation, if gH has finite order then there exists an n such that (gH)^n = (g^n)H = 1H. Thus, g^n1^{-1} \in H  \Rightarrow g^n \in H.

    (The bit where you go ` g^n1^{-1} \in H' holds because gH = hH \Leftrightarrow gh^{-1} \in H)
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    I see... Thank you

    So the contradiction is a consequence of the idea that ng is in A?

    I mean, if ng is in A, then there exists an m for which (ng)^m=1, means that mng=1, but I don't see the contradiction here :O
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    I see... Thank you

    So the contradiction is a consequence of the idea that ng is in A?

    I mean, if ng is in A, then there exists an m for which (ng)^m=1, means that mng=1, but I don't see the contradiction here :O
    What has just been proven is that if g+A has finite order then g \in A. Therefore, g+A=0+A. Apply this to what you are trying to prove.
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    Hmmm.. I still can't see the contradiction. Besides, how did you get to g+A, while we were discussing gn+A?
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    Hmmm.. I still can't see the contradiction. Besides, how did you get to g+A, while we were discussing gn+A?
    If g+A has finite order, n, then gn \in A and so gn has finite order (by the definition of A) and so there exists an m such that g(mn)=0. Thus, g has order mn, which is finite. Therefore, g \in A by the definition of A...and so g+A=0+A.
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    I guess I'm a having some trouble figuring this out :P
    I understand why what you've just said is correct, except that I don't understand why g+A=0+A, or how this leads to a contradiction.

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  10. #10
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    I guess I'm a having some trouble figuring this out :P
    I understand why what you've just said is correct, except that I don't understand why g+A=0+A, or how this leads to a contradiction.

    g+A=0+A as g is contained in A. It is contained in A because g must have finite order, and A is the group of all elements of finite order.
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    Quote Originally Posted by Swlabr View Post
    g+A=0+A as g is contained in A. It is contained in A because g must have finite order, and A is the group of all elements of finite order.
    Fine. Then everything is perfect. Where is the contradiction ?
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  12. #12
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    Fine. Then everything is perfect. Where is the contradiction ?
    Emm...I was assuming g+A \neq 0+A, but I suppose you don't have to.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by adam63 View Post
    Let G be a group, A be a group of all elements in G that have a finite order.

    Prove that the quotient group G/A contains no elements of finite order.

    Hmmm... If I assume that there is a gA in G/A for which there exists a natural n so that |gA|=n , then I get that : gA+...gA=gA (n times)

    How do I continue ? Am I doing this right?

    Thanks!
    I'd just like to remark that the set of all elements of G which have finite order is not in general a subgroup of G; for example, take the free product of two finite groups.

    In an abelian group, however, it is always a subgroup.
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    It's a little awkward, but I still don't understand it

    Bruno, what do you mean by the free product of two finite groups? AxB or AB?
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  15. #15
    MHF Contributor Swlabr's Avatar
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    What do you not understand? Bruno's comment, or the proof?

    Let S be the set of all elements of finite order in a group. Then it is possible to multiply two elements together to get an element of infinite order (for example, you can find 2x2 matrices which are of finite order multiplicatively, but when multiplied together they will form a matrix of infinite multiplicative order). This means that S is not necessarily a subgroup of your group. However, if S is a group, it will be normal, as conjugation preserves the order of an element (if g^n=1 then (h^{-1}gh)^n = h^{-1}g^nh=1).

    He also pointed out that in an abelian group, this set S will always be a group, as n(a+b)=1 where n= o(a)\cdoto(b).

    Do not worry about what a free-product is. You will learn that with time.
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