Is union of any number conjugacy classes is subgroup? Let $G$ be any group. We know that any subgroup $H$ of $G$ is normal if and only if $H$ is union of conjugacy classes. My first problem is let $a$ and $b$ are two elements of $G$. $Cl_G(a)$, $Cl_G(b)$, $Cl_G(identity)$ are conjugate classes of $a$ and $b$ and the identity of the group $G$ and $H$ is union of these classes, then is $H$ is subgroup of $G$? My second problem is if $|Cl_G(a)| + |Cl_G(b)| + 1 =n$ and $n$ divides $|G|$, then $H$ is definitely a normal subgroup? Or should we check that $H$ is subgroup or not then?
No. For example, consider $(\mathbb{Z}_4, +)$. Then $S=\{0, 1, 2\}$ is not a subgroup even though it is the union of conjugacy classes (and contains the identity). More generally, if $S\subset A$, $A$ abelian, is chosen such that $|A|&gt;|S|&gt;|A|/2$ then $S$ is not a subgroup.

As another counter-example, consider $(\mathbb{Z}, +)$. Then $S=\{-1, 0, 1\}$ is not a subgroup.

(These examples work because conjugacy classes in abelian groups are singletons. So more generally, in an abelian group every subset is a union of conjugacy classes.)
No. The conjugacy classes of $S_n$ are cycle types. Clearly $2$-cycles don't form a subgroup since they don't contain the identity.