Prove Centralizer Equality: n & k Relatively Prime

[tyrannosaurus]

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.

 

[rasmhop]

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 $C(a) \subseteq C(a^K)$ and $C(a^K) \subseteq C(a)$).

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]<br /> i.e. a^K and b commute. You then wish to show $bab^{-1}=a$. Here I have just stated the goal so we're clear about that.<br /> <br /> 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:<br /> $$(bab^{-1})^1 = (bab^{-1})^{xK+yn}$$<br /> so it suffices to show:<br /> $$(bab^{-1})^{xK+yn} = a$$<br /> See if you can do this. Remember that you have:<br /> $$ba^K b^{-1} = a^K \qquad a^n = 1$$[/tex]