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