I'm working on this abstract algebra problem that was given as homework to me, I'm in my first year of university and have a semester's background in basic abstract algebra. Here is the problem:
Let $G$ be a group. $\forall x \in G$ let $C(x)$ be $g\mid g \in G: gx=xg$.
Define the equivalence $\sim \forall x,y \in G$ if there exists $g \in G$ such that $g^{-1}xg=y$.
$\forall x \in G$, $[x]$ shall define the equivalence class of $x$. Let $D_x$ be the set $C(x)g\mid g\in G$.
Define $\psi: D_x \rightarrow [x]$ such that $\psi(C(x)g)=g^{-1}xg, \space \forall g \in G$.
Show that:
(a) $\psi$ is well defined.
(b) $\psi$ is a bijection.
(c) if $G$ is a finite group and $x \in G$, then $|[x]|$ divides $|G|$.
It's not clear to me how to demonstrate point a, that is, if $g_1,g_2 \in G$ such that $\psi(C(x)g_1) = \psi(C(x)g_2)$, then $C(x)g_1 = C(x)g_2$.
Also, in point b surjectivity is quite simple, but injectivity does not seem as intuitive.
Hint: (1). $C(x)g = C(x)h \Leftrightarrow gh^{-1} \in C(x) \Leftrightarrow gh^{-1}x = xgh^{-1} \Leftrightarrow h^{-1}xh = g^{-1}xg.$
(2). Injectivity follows from above. For surjectivity follows from the definition of $\sim.$
(3). Use Lagrange's Theorem.