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Math Help - Multiplicative group of a field.

  1. #1
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    Multiplicative group of a field.

    Hi:
    Let G be an abelian group (written multiplicatively), |G|=n. Denote by \mu (G) the product of all elements of G. Consider the subgroups G_2= \{x \in G: x^{2}=1\} and G^{2}=\{x^{2}: x \in G\}. Prove:
    (a) \mu(G)= \mu(G_2).
    (b) If G is a subgr of the multiplicative group of a field K and -1 \in G, then \mu(G)= -1 = (-1)^{n/2} \mu(G^{2}).

    Up to here, the statement of the problem. And it is in the statement, point (b), that I find a counter-example: For the statement seems to assert n is an even number. However, let K be finite, |K|= 2^{m} and take G to be the whole multiplicative group of K. Then |K| even and |G| is odd. Yet, 1, and therefore -1 belong to G, as the author of the problem requires (that is, -1 \in K and -1 not equal 0. So, -1 \in G).

    Am I right? Is the statement wrong? Any hint will be greatly appreciated. Good bye.
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  2. #2
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    Hi:
    Let G be an abelian group (written multiplicatively), |G|=n. Denote by \mu (G) the product of all elements of G. Consider the subgroups G_2= \{x \in G: x^{2}=1\} and G^{2}=\{x^{2}: x \in G\}. Prove:
    (a) \mu(G)= \mu(G_2).
    (b) If G is a subgr of the multiplicative group of a field K and -1 \in G, then \mu(G)= -1 = (-1)^{n/2} \mu(G^{2}).

    Up to here, the statement of the problem. And it is in the statement, point (b), that I find a counter-example: For the statement seems to assert n is an even number. However, let K be finite, |K|= 2^{m} and take G to be the whole multiplicative group of K. Then |K| even and |G| is odd. Yet, 1, and therefore -1 belong to G, as the author of the problem requires (that is, -1 \in K and -1 not equal 0. So, -1 \in G).

    Am I right? Is the statement wrong? Any hint will be greatly appreciated. Good bye.

    From where is this statement? If it is a from a book it may well a mistraslation and or a typo, and I think it should be \mu(G)= -1 = (-1)^{n/2\cdot\mu(G^{2})} , otherwise the very statment of the result is basically non-sensical: how can -1 equal -1 to some power times the number of elements in \mu(G^2)?? In the best of the cases this would mean \mu(G^2)=1 , which of course is false.
    Check this, I think it works fine with the correction.

    Tonio

    Pd. Si estás estudiando de un libro traducido ten cuidado: son comunes los errores de traducción y los tipográficos.
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  3. #3
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    It is from a German book on algebra, translated into English. I don't think it is in error. If char K not equal 2, then n is even. Not only this. In this case the statement is valid. I think I have the correct proof.

    You say \mu(G^{2})=1 is false. But consider K=\{0,1\}. In K, -1=1. So G=\{1\} and -1 \in G. As a result, G^{2}=\{1^{2}\} and \mu(G^{2})=1. Bear in mind: if G=\{a_1,a_2,...,a_k\} then, assuming the a_i^{2} are all diferent, \mu(G^{2})= a_1^{2}a_2^{2}...a_k^{2}
    Thank for your replay.
    Enrique.

    P.D.: estaba ayer contestando tu correo, cuando el sistema me dejo afuera. Decidi dejarlo para el dia siguiente. Aunque ya es seguro que nadie conteste este.
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  4. #4
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    It is from a German book on algebra, translated into English. I don't think it is in error. If char K not equal 2, then n is even. Not only this. In this case the statement is valid. I think I have the correct proof.

    You say \mu(G^{2})=1 is false.

    I meant it was false in a general context, but I misread: I though \mu(G) was the number of elements in G and etc. My bad.

    Tonio

    But consider K=\{0,1\}. In K, -1=1. So G=\{1\} and -1 \in G. As a result, G^{2}=\{1^{2}\} and \mu(G^{2})=1. Bear in mind: if G=\{a_1,a_2,...,a_k\} then, assuming the a_i^{2} are all diferent, \mu(G^{2})= a_1^{2}a_2^{2}...a_k^{2}
    Thank for your replay.
    Enrique.

    P.D.: estaba ayer contestando tu correo, cuando el sistema me dejo afuera. Decidi dejarlo para el dia siguiente. Aunque ya es seguro que nadie conteste este.
    .
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