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).

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- Sep 24th 2008, 08:04 PMbaseballmathShow 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 PMThePerfectHacker
- Sep 24th 2008, 09:33 PMbaseballmath
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 AMThePerfectHacker
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 $\displaystyle x\in C(a^{-1})$.