Conjugacy Classes, Normal Subgroups and Guaranteed Existence Suppose we have a group $G$ of order $24$, and we know apriori that this group has $5$ conjugacy classes, or sizes $1$, $3$, $6$, $6$, and $8$.
The question I'm presented with is whether or not one can conclude that $G$ has exactly two normal subgroups.
I know that normal subgroups are the union of conjugacy classes. Moreover, this union must include the singleton conjugacy class for the identity, and the order of the union should divide the order of the group. It seems plausible in this case then that we could take the conjugacy classes and obtain two normal subgroups of order $1 + 3 = 4$ and $1 + 3 + 8 = 12$. But is this guaranteed?
 A: If you take the classification of groups of order 24, you find that there is only one group with 5 conjugacy classes (namely $S_4$) and it has exactly two proper normal subgroups. So (somewhat casuistic) the naive answer is "Yes".
But the existence of normal subgroups does not follow just from the class sizes. It is not hard to come up with examples of groups and class sizes, where a union of classes satisfies the conditions, but does not form a normal subgroup. So as soon as you ask the same question in the generic case, the answer is "No".
A: You could argue that $G$ has at most $2$ proper non-trivial normal subgroups since the combinations you found for writing the size of a normal subgroup as a sum of class sizes are exhaustive.
But a group of size $24$ is not simple, so one of the two possibilities must occur. If $G$ has a normal subgroup of order $4$, say $N$, look at the quotient $G/N$ which has order $6$. By Sylow (or Cauchy), $G/N$ has a normal subgroup $H/N$ of order $3$ and thus $H$ is normal in $G$ and of order $12$. A similar argument will allow you to deduce the existence of a normal subgroup of order $4$ given the existence of a normal subgroup of order $12$.
In general, however, as pointed out in the other answer it does not follow that if the sum of some class sizes is a divisor of the order of the group then a normal subgroup of that order must exist.
