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Math Help - prove S is subgroup of G where S is stated with some conditions

  1. #1
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    prove S is subgroup of G where S is stated with some conditions

    Hi

    here is the problem i am trying to do.

    let G be a finite group and let S be a nonempty subset of G.
    suppose  \mathit{e} \in S (e is an identity element)

    and that S is closed with respect to multiplication.Prove that S

    is a subgroup of G.

    This is my attempt at solution.

    So since  \mathit{e} \in S, to show that S is subgroup

    of G, we
    need to show that \forall x \in S \;\; x^{-1} \in S

    Since G is finite, S is finite too. Let

    S=\{\mathit{e},a_1,a_2 \cdots ,x,\cdots a_n \}

    where  x is an arbitrary element of the S. So we need to
    show that x^{-1} \in S. lets multiply each element of
    S by x. So we have set P

    P=\{x\ast \mathit{e},x\ast a_1 \cdots ,x \ast x \cdots ,x \ast a_n \}

    since S is closed under multiplication , all elements of P are just
    rearranged members of S. So S=P

    By our contruction of S ,

    x\ne \mathit{e} \;\Rightarrow \;x\ast \mathit{e} \ne \mathit{e}

    But  \mathit{e} \in S

    So \exists y \in S \;\; (x\ast y =\mathit{e})

    \Rightarrow y = x^{-1}

    since x is arbitrary ,

    \forall x \in S \;\; x^{-1} \in S

    So S is closed with respect to inverses.

    So S is a subgroup of G.


    Is it correct .... my reasoning ?

    thanks
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Re: prove S is subgroup of G where S is stated with some conditions

    No - you do not need to prove that S is closed with respect to inverses, you need to prove that S contains inverses - that x^{-1}\in S is x\in S. To do this, note that x has finite order. So...

    Also, it is interesting to note that this theorem does not hold for infinite groups. Can you come up with a counter-example?
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  3. #3
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    Re: prove S is subgroup of G where S is stated with some conditions

    Hi Swlabr

    Yes I used wrong wording. But I did prove that  x^{-1} \in S . is it wrong ?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Re: prove S is subgroup of G where S is stated with some conditions

    Quote Originally Posted by issacnewton View Post
    Hi Swlabr

    Yes I used wrong wording. But I did prove that  x^{-1} \in S . is it wrong ?
    Ah-sorry-your problem is that you need to show that multiplying by x is a bijection, which you haven't done. To do this, it is sufficient to prove that it is an injection. That is, if xg=xh then g=h. Can you see how you would do this?
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  5. #5
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    Re: prove S is subgroup of G where S is stated with some conditions

    if xg=xh \Rightarrow x^{-1}xg=x^{-1}xh \Rightarrow g=h , right ? so do I prove injection from S to P ?
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Re: prove S is subgroup of G where S is stated with some conditions

    Quote Originally Posted by issacnewton View Post
    if xg=xh \Rightarrow x^{-1}xg=x^{-1}xh \Rightarrow g=h , right ? so do I prove injection from S to P ?
    Yes, the line you just posted proved that this is an injection. Which completes your proof.

    Now, for the sake of completeness, you should work out where you used the fact that the group was finite.
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  7. #7
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    Re: prove S is subgroup of G where S is stated with some conditions

    I think when I declared the set S , I only took finite elements. Was that the place I used that fact ?
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Re: prove S is subgroup of G where S is stated with some conditions

    Quote Originally Posted by issacnewton View Post
    I think when I declared the set S , I only took finite elements. Was that the place I used that fact ?
    The finite set bit is used when you are trying to show that there is a bijection between S and P. Can you explain why this would be?
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  9. #9
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    Re: prove S is subgroup of G where S is stated with some conditions

    its because , for the set to be finite , there has to be a injection from S to \mathbb{N}
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Re: prove S is subgroup of G where S is stated with some conditions

    Quote Originally Posted by issacnewton View Post
    its because , for the set to be finite , there has to be a injection from S to \mathbb{N}
    No. You have a map going from S to P, f: S\rightarrow P. It is surjective, by definition, and you have proven it is injective. Further, P\subset S. Now, a finite group has no proper subsets which have the same cardinality, but infinite sets can! Thus, S=P because S is finite.

    Now, can you come up with an infinite counter-example?
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  11. #11
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    Re: prove S is subgroup of G where S is stated with some conditions

    counterexample to what ?
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  12. #12
    MHF Contributor Swlabr's Avatar
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    Re: prove S is subgroup of G where S is stated with some conditions

    Quote Originally Posted by issacnewton View Post
    counterexample to what ?
    A group which contains a set which is closed under multiplication but not inverses.
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  13. #13
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    Re: prove S is subgroup of G where S is stated with some conditions

    I think \mathbb{Z} is a group under addition. But \mathbb{Z}^{+}\cup \{0\} would be subset of \mathbb{Z} but the inverses
    are not in \mathbb{Z}^{+}\cup \{0\}
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  14. #14
    MHF Contributor Swlabr's Avatar
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    Re: prove S is subgroup of G where S is stated with some conditions

    Quote Originally Posted by issacnewton View Post
    I think \mathbb{Z} is a group under addition. But \mathbb{Z}^{+}\cup \{0\} would be subset of \mathbb{Z} but the inverses
    are not in \mathbb{Z}^{+}\cup \{0\}
    Yeah, that's a good example. You could also take the integers under multiplication, which is a subset of the rationals under multiplication but obviously not a subgroup.
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