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Math Help - Order of groups involving conjugates and abelian groups

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    Order of groups involving conjugates and abelian groups

    1. Let G be a group and let a be in G. An element b in G is called a conjugate of a if there exists an element x in G such that b = xax^-1. Show that any conjugate of a has the same order as a.

    2. Let G be an abelian group and let x,y be in G. Suppose that x and y are of finite order. Show that xy is of finite order and that, in fact, o(xy) (order of xy) divides o(x)o(y).
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    Quote Originally Posted by Janu42 View Post
    1. Let G be a group and let a be in G. An element b in G is called a conjugate of a if there exists an element x in G such that b = xax^-1. Show that any conjugate of a has the same order as a.
    Hint: If b = xax^{-1} then b^n = xa^n x^{-1}.

    2. Let G be an abelian group and let x,y be in G. Suppose that x and y are of finite order. Show that xy is of finite order and that, in fact, o(xy) (order of xy) divides o(x)o(y).
    Note (ab)^k = a^kb^k since G is abelian.
    If |x|=n,|y|=m then (xy)^{nm} = (x^n)^m(y^m)^n = 1.
    Thus, |xy| \text{ divides }|x||y|.
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    Thanks again. Also, quick one:

    G = {1,2,3,4,5,6}
    The group (G, *), where * is multiplication mod 7. For example, 2*4 = 8 = 1, since 8 = 1 mod 7.

    How do I show it's cyclic? I know it is, but which element of G can I use as the generator?

    If I do <1>, doesn't that mean when I do 1^2 it is 1*1 and 1^3 is 1*1*1?
    Because then I would have 1 every time.
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    Quote Originally Posted by Janu42 View Post
    Thanks again. Also, quick one:

    G = {1,2,3,4,5,6}
    The group (G, *), where * is multiplication mod 7. For example, 2*4 = 8 = 1, since 8 = 1 mod 7.

    How do I show it's cyclic? I know it is, but which element of G can I use as the generator?

    If I do <1>, doesn't that mean when I do 1^2 it is 1*1 and 1^3 is 1*1*1?
    Because then I would have 1 every time.
    Find an element a so that G = \{ a,a^2,a^3,a^4,a^5,a^6\}, remembmer a^k means a raised to k reduced mod 7.
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    OK, one more, I'm pretty sure this is simple. I'm pretty sure it has something to do with the order being a multiple of 2.

    Show that if the order of G is an even integer, then there is an element x in G such that x is not equal to e, and x^2 = e.
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    Quote Originally Posted by Janu42 View Post
    OK, one more, I'm pretty sure this is simple. I'm pretty sure it has something to do with the order being a multiple of 2.

    Show that if the order of G is an even integer, then there is an element x in G such that x is not equal to e, and x^2 = e.
    Assume not. Then for every x\not = e there is y so that xy=e. Thus, all elements \not = e can be paired with their inverses. But then that means there are odd number of elements.
    A contradiction.
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