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Thread: 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 $\displaystyle a \in G$ and say $\displaystyle n = |G|$ consider $\displaystyle a,a^2,...,a^n$. Now if $\displaystyle a^i = a^j $ for some $\displaystyle i>j$ then $\displaystyle a^{i-j} = e$ where $\displaystyle 1\leq i-j\leq n$. Otherwise if $\displaystyle a,a^2,...,a^n$ are all distinct then by pigeonholing it is a permutation of element $\displaystyle G$ and so $\displaystyle a^k = e$ for $\displaystyle 1\leq k\leq n$.

    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 $\displaystyle H_a$ is a subgroup of G. What have you done? Show what you tried to do.
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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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    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 $\displaystyle a$. And it is $\displaystyle H_a = \{ x\in G| xa = ax\}$. So basically it is the set of all $\displaystyle x$ that commute with all $\displaystyle a$. Let $\displaystyle x\in H_a$ then $\displaystyle ax = xa$ so $\displaystyle y(ax) = y(xa)$ but $\displaystyle y\in H_a$ so $\displaystyle ay=ya$ this means $\displaystyle a(yx) = (yx)a$ so $\displaystyle xy\in H_a$. This $\displaystyle H_a$ has the identity $\displaystyle e$. Now it has the associative property because G which contains it has that property. Finally if $\displaystyle x\in H_a$ then it means $\displaystyle ax = xa$ thus $\displaystyle axa^{-1} = xaa^{-1} \implies axa^{-1} = x$ thus $\displaystyle a^{-1} axa^{-1} = a^{-1} x\implies xa^{-1} = a^{-1} x$ and so $\displaystyle a^{-1} \in H_a$. So this proves it is a subgroup.
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