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Centralizer proof

  • #1

Homework Statement


If a is an element of a group and |a|=n, prove that C(a)=C(a^k) when k is relatively prime to n.


Homework Equations


If n and k are relatively prime, then there exists integers s and t such that ns+kt=1.
The centralizer a in G, C(a) is the set of all elements in a group G that commute with a. C(a)={g an element of G|ga-ag}


The Attempt at a Solution


I tried proving it by double containment, but I couldn't show that C(a^K) is contained in C(a). Should I try a proof by contradiction? I will be grateful for any help.
 

Answers and Replies

  • #2
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3

The Attempt at a Solution


I tried proving it by double containment, but I couldn't show that C(a^K) is contained in C(a).
I think double containment is the easiest method (I have never heard the term double containment, but I'm assuming it refers to proving [itex]C(a) \subseteq C(a^K)[/itex] and [itex]C(a^K) \subseteq C(a)[/itex]).

You want to show that if b is in C(a^K), then b is in C(a). In other words assume:
[tex]ba^Kb^{-1} = a^K[/itex]
i.e. a^K and b commute. You then wish to show [itex]bab^{-1}=a[/itex]. Here I have just stated the goal so we're clear about that.

To say that K and n are relatively prime is equivalent to saying that there exists integers x, y such that xK+yn=1. Now you have:
[tex](bab^{-1})^1 = (bab^{-1})^{xK+yn}[/tex]
so it suffices to show:
[tex](bab^{-1})^{xK+yn} = a[/tex]
See if you can do this. Remember that you have:
[tex]ba^K b^{-1} = a^K \qquad a^n = 1[/tex]
 

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