Conjugacy classes and Centers

Question

I need to show that:

The conjugacy class of $a$ contains $a$ (and no other element) iff $a$ is in the center.

Here's how I attempted the proof:

$(\Rightarrow )$ Suppose $\text{cl}(a)=\{ a\}$. Then for every $x\in G$, $xax^{-1}=a$. Thus, $a$ is the center of $G$.

$(\Leftarrow )$ Assume $a$ is the center of G. Then $a\in \text{cl}(a)$ since $xax^{-1}=a$ for every $x \in G$. Let $b$ be any element in $\text{cl}(a)$. Then $b=yay^{-1}$ for some $y\in G$. But $b=yay^{-1} \Rightarrow by=ya=ay$. Thus, by the cancellation property, $b=a$. Hence, $\text{cl}(a)=\{ a\}$.

Is my proof correct? Do I need to fix it somewhere?

Answer 1

Your proof is correct but you have some unnecessary steps at the end. You don't have to prove $b=a$ the way you did: you already know that $xax^{-1}=a$ for all $x$, and so in particular you can take $x=y$ to conclude immediately $b=yay^{-1}=a$.

Answer 2

It's correct, but it could be easier : Notice that elements of the center are elements that commute with all elements of the group. Therefore, $a$ is in the center $\iff$ $a$ commute with all elements of $G$ $\iff$ $cl(a)=\{a\}$ (almost by definition).

Answer 3

since by definition $cl(a) = \{x^{-1}ax:x \in G\}$ we have: $$ a \in Z(G) \iff \forall x \in G.ax=xa \iff \forall x \in G.x^{-1}ax=a \iff \{x^{-1}ax:x \in G\} = \{a\} $$