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Math Help - orders of groups

  1. #1
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    orders of groups

    Let x and y be elements of a group G and m and n coprime positive integers. Show that if x^m = y^m and x^n = y^n then x=y

    Let x and y be elements of a group G of orders m and n respectively. Show that if x and y commute and m and n are coprime, then order(xy) = mn


    not sure at all on this

    many thanks
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  2. #2
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    If m and n are coprime then there exist integers x and y such that

    xm+yn = 1

    Can you see where to take this?
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  3. #3
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    not really.. could you explain a bit more please?
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  4. #4
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    Oh, I choose poor letters for that, I meant there are integers a and b such that

    am + bn = 1
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  5. #5
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    ok.. so we can raise x and y to each of those expressions...

    so x^(am).x^(bn) = x
    y^(am).y^(bn)=y

    since x^m = y^m, x^am = y^am..

    so y^am.x^bn = x
    hence x = y

    or am I waayyyy off track?

    thankyou!
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  6. #6
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    Nope; nail on the head
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  7. #7
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    thankyou!
    Any ideas on the next bit (I tried to do it without being able to do the first part - just by assuming the result but still couldnt get anywhere with it.. we've not covered lagrange's theorem)
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  8. #8
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    Can you show that (xy)^(mn) = 1?

    Can you show that there is nothing smaller which satisfies?
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  9. #9
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    I can show that, yes.. to show nothing smaller satisfies would it be sufficient to say, suppose there exists ab < mn such that

    xy^ab = 1

    then x^ab.y^ab = 1

    => x has order ab.. contradiction?
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  10. #10
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    Nope. y^ab could be (x^{ab})^{-1}
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  11. #11
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    on the subject of groups.. could you please help me with another questipon.. about giving examples..

    of a Group G and a proper subgroup H of G such that:
    1) G is infinite and H is finite
    2) G is infinite and non-abelian and H is abelian
    3) G is infinite and abelian and H is infinite
    4) G is infinite and H is not-abelian..

    I'm really struggling..

    for 1 could we have

    G = the set of rotations about 0, H = the set of rotations of 90 degress for example?
    and simililarly could we have that for 4?
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  12. #12
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    ahh ok.. so is contradiction the right way to go about it?
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  13. #13
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    Some solutions

    I dunno if you can use trivial subgroups or not, but that makes some really easy.

    1) G=\mathbb{Z} and H={0}=e
    Another one that isn't as cheap could be polynomials under addition of a finite field, like Z_p and just using H as the subgroup of polynomials with of degree 0 (the constant terms in the field).

    2) G=GL(n,\mathbb{R}) invertible n by n matrices with entries from the reals under multiplication. If you want to be lame, let H be the identity matrix. Let H be the set matrices that have the same real number on the diagonal and 0s elsewhere. H={A \in G|A=rI_n, r \in \mathbb{R}} Note I_n is the identity matrix.

    3)This one is pretty easy, tons of examples \mathbb{C} \supset \mathbb{R}\supset \mathbb{Q} \supset \mathbb{Z} All of these groups are infinite and abelian and the containment is proper, take your pick.

    4) You just need some infinite nonableian groups. G=GL(n,\mathbb{R}) would be a good start. Then like for subgroups upper triangular matrices might be a good one.

    Well that is a start, you should try to think about some more on your own for sure, but this is the idea.
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