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Math Help - Prove for cyclic group

  1. #1
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    Prove for cyclic group

    I cant find answer to this:

    If G is finite group of order n, G is cyclic iff there is element in G with order n, that is a^n = e (the identity in G for the operation in G).

    i dont know where to start, i thought of taking it this way but im afraid that's not the answer, or is it?..

    if a is any element in G, and G is of order n, and you get this element a to power n for any n E Z, and if you get this a^n = e, then you can do this..

    a^n = e => a^n-1 = a^-1, so the inverse of a is a^n-1, and you can go on like this and find all the inverses for a^k, for all k =1,..,n,
    since you got a^n, you can find a^1, and you can do that like this:
    a^n = e => a^n+1 = a^1, and you can go on like this and get all a^k for k=1,..,n so you have all the elements and their inverses in G making a the generator of G (G = <a>).

    But im afraid this is not quite true cause i cant tell if a is the generator for sure , i lack of confidence to tell that, so if someone can guide me here thanks!!
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  2. #2
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    Quote Originally Posted by goroner View Post
    I cant find answer to this:

    If G is finite group of order n, G is cyclic iff there is element in G with order n, that is a^n = e (the identity in G for the operation in G).


    This is absurd: in a group of order n we have that x^n=e for ANY element in the group, and the

    group may perfectly well not be cyclic. What you probably meant is that there's an element x

    in the group s.t. x^n =e AND x^m\neq e\,\,\,\forall m<n\,,\,\,m,n\in\mathbb{N}

    And this is simply the definition of cyclic group...

    Tonio


    i dont know where to start, i thought of taking it this way but im afraid that's not the answer, or is it?..

    if a is any element in G, and G is of order n, and you get this element a to power n for any n E Z, and if you get this a^n = e, then you can do this..

    a^n = e => a^n-1 = a^-1, so the inverse of a is a^n-1, and you can go on like this and find all the inverses for a^k, for all k =1,..,n,
    since you got a^n, you can find a^1, and you can do that like this:
    a^n = e => a^n+1 = a^1, and you can go on like this and get all a^k for k=1,..,n so you have all the elements and their inverses in G making a the generator of G (G = <a>).

    But im afraid this is not quite true cause i cant tell if a is the generator for sure , i lack of confidence to tell that, so if someone can guide me here thanks!!
    .
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  3. #3
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    i didnt mean what you wrote, i just try to solve task in the exercise part of the book im reading and this is what is written there:

    Prove that:
    "If G is group of order n, G is cyclic iff G has an element of order n"
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  4. #4
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    we have to prove TWO things:

    1.) if |G| = n and G has an element (let's call it g) of order n, then G is cyclic.

    2.) If |G| = n and G is cyclic, G has an element of order n.

    the second one is easier to prove, so we'll start there. we know G is cyclic. so G has an element x, with G = <x> = {x, x^2, x^3,.....,x^k,e}

    as you can see, we have k powers of x, and the identity e. since these are ALL the elements of G, we have: k+1 = n.

    this means that k = n-1. thus x, x^2, x^3,....,x^(n-1) are all NOT the identity. but we know that x^n IS the identity

    (since this is true of EVERY element of a group of order n). so x is the element of order n we are looking for.

    now, let's see about #1. here, we only know that G has an element g of order n. this means:

    g^k ≠ e for 0 < k < n, and g^n = e. so let's consider THIS subgroup of G: <g>. since g is of order n,

    {g,g^2,g^3,....,g^(n-1), g^n = e} are all different. why?

    suppose g^k = g^m for 0 < k,m < n, with k ≠ m. we can assume k > m (or else just switch them around).

    then (g^k)(g^(-m)) = (g^m)(g^(-m)), so

    g^(k-m) = e. but k > m, so k-m > 0. and k < n, so k-m < k < n. and this would mean that k-m is a smaller (positive) power of g

    equal to the identity than n. but n is the order of g, so this cannot be the case (n is the smallest positive power of g equal to e).

    since this is a contradiction, we can't possibly have g^k = g^m for 0 < k,m < n and k ≠ m. so the powers of g:

    {g, g^2, g^3,.....,g^(n-1), g^n = e} are all distinct. but this is n elements, which means that <g> is all of G. so g generates G,

    thus G is cyclic.
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  5. #5
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    Quote Originally Posted by goroner View Post
    i didnt mean what you wrote, i just try to solve task in the exercise part of the book im reading and this is what is written there:

    Prove that:
    "If G is group of order n, G is cyclic iff G has an element of order n"

    Well, then it all boils down to what's the definition of "cyclic group" in that book. Apparently it is that a group G is

    cyclic if there's an element g\in G\,\,s.t.\,\,\forall x\in G\,,\,\,x = g^m for some integer m.

    The trick: if g\in G has order n (this means what I wrote and NOT what you wrote) then count the elements

    in the set \{g^0,g,g^2,...,g^{n-1}\}

    Tonio
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  6. #6
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    thanks a lot guys =)
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