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

1. ## 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).

2. 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 $x\in C(a^{-1})$ then $x\in C(a)$
2)Show if $x\in C(a)$ then $x\in C(a^{-1})$

3. 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...

4. I do one direction.
Let $x\in C(a)$.
Therefore, $x^{-1} \in C(a)$ because $C(a)$ is a subgroup.
By definition $ax^{-1} = x^{-1}a$.
Thus, $(ax^{-1})^{-1} = (x^{-1}a)^{-1}$.
Thus, $xa^{-1} = a^{-1}x$.
By definition $x\in C(a^{-1})$.