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Math Help - Group Theory

  1. #1
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    Group Theory

    I have the following problem which i have no idea how to proceed.

    Let G be a finite group with odd number of elements. Show that for every g \in G, there exists an element h \in G such that g=h^2.

    I have tried to construct a subgroup H of G defined by H=\{h\in G \mid g=h^2,   \forall g \in G \}. However, I cant proceed on after that. Perhaps someone can guide me on how to proceed on with this problem. Thank You.
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  2. #2
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    Re: Group Theory

    hint: use lagrange's theorem, to show that the order of every element of G is odd. now show that <g> = <g^2> for every g in G.

    why can we use this to find h?
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  3. #3
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    Re: Group Theory

    let me try the solution with the help of ur hint.

    Let K = <g>. Since K is cyclic, the order of K = order of g. Also, K is a subgroup of G. By lagrange theorem, \left\vert K \right\vert divides \left\vert G \right\vert. Since \left\vert G \right\vert is odd, therefore \left\vert K \right\vert is also odd. Hence, order of any elements in G is odd.

    Next to show: < g> = < g^2>.
    For this part i am not very sure but i will try:
    <g> = { 1, g, g^2, g^3, g^4, ...} and < g^2> = { 1, g^2, g^4, g^6, ...}. Since order of g is odd, therefore, g^(2m+1) = 1, where 1 is the identity of G. Hence, <g> reduces to <g> = { 1, g^2, g^4, ...}. Thus <g> = <g^2>.

    Is the above reasoning correct?
    However I dont really know how this allows us to find h.
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  4. #4
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    Re: Group Theory

    for the second part, since the powers of g^2 "go up by 2's", and since |G| = 2m+1, when we get to g^(2m+2), we wind up back at g.

    now surely <g^2> is a subgroup of <g> (every even power of g is also a power of g).

    but since g is in <g^2>, <g> must be a subgroup of <g^2>, so.... (fill in the blank).

    but this means that g = g^(2m+2) = [g^(m+1)]^2, so what do you think we should choose for h?
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  5. #5
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    Re: Group Theory

    Choose h= g^(m+1) ?
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  6. #6
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    Re: Group Theory

    and does h^2 = g then? if so, you're done.
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