TITLE: Prove that $ a\in Z(G) $
QUESTION [5 upvotes]: Let $ (G,\cdot ) $ be a group with $ |G|=2m+1,m\in \mathbb{N} $ and $ a\in G $ so that there exists $ n\in \mathbb{N} $ with $ a^{n}\cdot x=x\cdot a,\forall x\in G\setminus A $, where $ A=\left \{ a^{k}|k\in \mathbb{Z} \right \}. $ 
Prove that $ a\cdot x=x\cdot a,\forall x\in G. $
Obviously, if $ x\in A $, $ a\cdot x=x\cdot a. $ 
If $ x\notin A $, the only thing I've found is that $ a^{n}\cdot (x\cdot a^{n-1})^{2m}\cdot x\cdot a^{-1}=e $.

REPLY [2 votes]: Note that $x \in G\setminus A$ if and only if $x^{-1} \in G\setminus A$. Hence, the property implies that $x^{-1}ax=a^n=xax^{-1}$. So, $x^2a=ax^2$ and also $x^{-2}a=ax^{-2}$. But $x$ has odd order, hence $x^{2k+1}=1$ for some positive integer $k$. But then $xa=x^{-2k}a=(x^2)^{-k}a=a(x^2)^{-k}=ax^{-2k}=ax$.