The set of all conjugation classes of group $G$ form a partition of $G$

Solution 1:

If $g\in G$, then $g$ belongs to the conjugation class $[g]$ of $g$. Therefore, $G=\bigcup_{g\in G}[g]$.

And if $[g_1]\cap[g_2]\neq\emptyset$, take $g\in[g_1]\cap[g_2]$. Then there are $h_1,h_2\in G$ such that $g=h_1g_1{h_1}^{-1}$ and that $g=h_2g_2{h_2}^{-1}$. But then\begin{align}g_1&={h_1}^{-1}gh_1\\&={h_1}^{-1}{h_2}g_2{h_2}^{-1}h_1\\&=({h_2}^{-1}h_1)^{-1}g_2({h_2}^{-1}h_1)\end{align}and so $[g_1]=[g_2]$.

Solution 2:

Hint:

You can prove this by showing that the relation $\sim$ on $G$ defined by:$$g\sim h\iff \exists x\in G\;[gx=xh]$$is an equivalence relation.

The equivalence classes (they form a partition of $G$) are exactly the conjugation classes.