Show centralizer of a = centralizer of a^-1

**baseballmath** · September 24th 2008, 08:04 PM

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).

**ThePerfectHacker** · September 24th 2008, 09:04 PM

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 $x\in C(a^{-1})$

**baseballmath** · September 24th 2008, 09:33 PM

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...

**ThePerfectHacker** · September 25th 2008, 06:47 AM

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 $x\in C(a^{-1})$ .

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