= Solution
{c}
The <orbit> of $x$ under <conjugation> is its <conjugacy class>
$$
C(x)=\{gxg^{-1}:g\in G\}.
$$
It contains $x$ by using $g=e$, so the <conjugacy classes> cover $G$. If $z=gxg^{-1}$, then $x=g^{-1}zg$, and the <group action> identity shows that conjugating either element further gives the same <orbit>; hence $C(z)=C(x)$. If $C(x)$ and $C(y)$ intersect at $z$, this gives $C(x)=C(z)=C(y)$. Thus distinct <conjugacy classes> are disjoint and \b[the <conjugacy classes> form a <partition of a set> $G$].
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