# Show centralizer of a = centralizer of a^-1

• Sep 24th 2008, 08:04 PM
baseballmath
Show centralizer of a = centralizer of a^-1
Help. I feel like i'm over complicating this in my head.

Let G be a group, and let a be in G. Prove that C(a)=C(a^-1).
• Sep 24th 2008, 09:04 PM
ThePerfectHacker
Quote:

Originally Posted by baseballmath
Help. I feel like i'm over complicating this in my head.

Let G be a group, and let a be in G. Prove that C(a)=C(a^-1).

1)Show if $\displaystyle x\in C(a^{-1})$ then $\displaystyle x\in C(a)$
2)Show if $\displaystyle x\in C(a)$ then $\displaystyle x\in C(a^{-1})$
• Sep 24th 2008, 09:33 PM
baseballmath
I'm struggling with getting to show that. It's given that ag=ga and a^1g=ga^1.

I tried to go with
a^1 g a
= g a^1 a
= g e

and: a g a^1
= g a a^1
= g e
so a a^1 = a^1 a But I feel I'm making assumptions somewhere so the proof doesn't hold. I'm so lost! Thanks for your reply, it makes sense, I'm just not sure how to show what you're saying...
• Sep 25th 2008, 06:47 AM
ThePerfectHacker
I do one direction.
Let $\displaystyle x\in C(a)$.
Therefore, $\displaystyle x^{-1} \in C(a)$ because $\displaystyle C(a)$ is a subgroup.
By definition $\displaystyle ax^{-1} = x^{-1}a$.
Thus, $\displaystyle (ax^{-1})^{-1} = (x^{-1}a)^{-1}$.
Thus, $\displaystyle xa^{-1} = a^{-1}x$.
By definition $\displaystyle x\in C(a^{-1})$.