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Thread: Group Problem

  1. #1
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    Group Problem

    Let $\displaystyle G$ be a group. Suppose that
    $\displaystyle (\alpha\ast\beta)^3=\alpha^3\ast\beta^3$ -----(1)
    for all $\displaystyle \alpha, \beta \in G$.
    (i)Show that for all $\displaystyle \alpha, \beta \in G$, then
    $\displaystyle (\beta\ast\alpha)^2=\alpha^2\ast\beta^2$ and $\displaystyle \alpha^3\ast\beta^2=\beta^2\ast\alpha^3$.
    (ii)If $\displaystyle G$ satisfies (1) and the condition that the identity element $\displaystyle e$ in $\displaystyle G$, is the only element in $\displaystyle G$ satisfying $\displaystyle x^3=e$, show that for all $\displaystyle \alpha, \beta \in G$,
    $\displaystyle \alpha^2\ast\beta=\beta\ast\alpha^2$
    and conclude that $\displaystyle G$ is abelian.
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  2. #2
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    Proving part (i) is okay. The difficult portion is part (ii).
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  3. #3
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    Quote Originally Posted by Markeur View Post
    Let $\displaystyle G$ be a group. Suppose that
    $\displaystyle (\alpha\ast\beta)^3=\alpha^3\ast\beta^3$ -----(1)
    for all $\displaystyle \alpha, \beta \in G$.
    (i)Show that for all $\displaystyle \alpha, \beta \in G$, then
    $\displaystyle (\beta\ast\alpha)^2=\alpha^2\ast\beta^2$ and $\displaystyle \alpha^3\ast\beta^2=\beta^2\ast\alpha^3$.
    (ii)If $\displaystyle G$ satisfies (1) and the condition that the identity element $\displaystyle e$ in $\displaystyle G$, is the only element in $\displaystyle G$ satisfying $\displaystyle x^3=e$, show that for all $\displaystyle \alpha, \beta \in G$,
    $\displaystyle \alpha^2\ast\beta=\beta\ast\alpha^2$
    and conclude that $\displaystyle G$ is abelian.
    Hint 1. By condition (1), the map given by $\displaystyle f(\alpha) = \alpha^3$ is a homomorphism from the group to itself.

    Hint 2. If the kernel of $\displaystyle f$ consists solely of the identity, then $\displaystyle f$ is injective. In particular, if $\displaystyle (\alpha^2\ast\beta)^3 = (\beta\ast\alpha^2)^3$ then $\displaystyle \alpha^2\ast\beta=\beta\ast\alpha^2$.

    Hint 3. Use the properties in (i) to show that $\displaystyle (\alpha^2\ast\beta)^3 = (\beta\ast\alpha^2)^3$.

    To conclude that G is abelian, notice that $\displaystyle (\alpha\ast\beta)^3 = \alpha^3\ast\beta^3 = \alpha^2\ast\beta^2\ast\alpha\ast\beta$, from which $\displaystyle (\alpha\ast\beta)^2 = \alpha^2\ast\beta^2$, and it easily follows that $\displaystyle \beta\ast\alpha = \alpha\ast\beta$.
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