# Abstract: More groups

• Feb 10th 2008, 03:50 PM
fulltwist8
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}
• Feb 10th 2008, 03:54 PM
ThePerfectHacker
Quote:

Originally Posted by fulltwist8
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 $1\leq k\leq n$.

Quote:

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.
• Feb 10th 2008, 03:58 PM
fulltwist8
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?
• Feb 10th 2008, 04:08 PM
ThePerfectHacker
Quote:

Originally Posted by fulltwist8
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.