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Math Help - Abstract: More groups

  1. #1
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    Abstract: More groups

    1. Show that if a is in G, where G is a finite group with identity e, then there exists n in positive integers such that a^n=e


    2. Let G be a group and let a be one fixed element of G. Show that H (sub a)= {x in G: xa=ax}
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  2. #2
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    Quote Originally Posted by fulltwist8 View Post
    1. Show that if a is in G, where G is a finite group with identity e, then there exists n in positive integers such that a^n=e
    Let a \in G and say n = |G| consider a,a^2,...,a^n. Now if a^i = a^j for some i>j then a^{i-j} = e where 1\leq i-j\leq n. Otherwise if a,a^2,...,a^n are all distinct then by pigeonholing it is a permutation of element G and so a^k = e for 1\leq k\leq n.

    2. Let G be a group and let a be one fixed element of G. Show that H (sub a)= {x in G: xa=ax}
    I assume you want to show H_a is a subgroup of G. What have you done? Show what you tried to do.
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  3. #3
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    yes, prove that it's a subgroup. sorry, i forgot that part! i have no idea where to start on number two. is it like saying x*a = a*x?
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  4. #4
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    Quote Originally Posted by fulltwist8 View Post
    yes, prove that it's a subgroup. sorry, i forgot that part! i have no idea where to start on number two. is it like saying x*a = a*x?
    First, this is subgroup has a name. It is called the centralizer of a. And it is H_a = \{ x\in G| xa = ax\}. So basically it is the set of all x that commute with all a. Let x\in H_a then ax = xa so y(ax) = y(xa) but y\in H_a so ay=ya this means a(yx) = (yx)a so xy\in H_a. This H_a has the identity e. Now it has the associative property because G which contains it has that property. Finally if x\in H_a then it means ax = xa thus axa^{-1} = xaa^{-1} \implies axa^{-1} = x thus a^{-1} axa^{-1} = a^{-1} x\implies xa^{-1} = a^{-1} x and so a^{-1} \in H_a. So this proves it is a subgroup.
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