# 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?

• Do you count the group and the trivial group among the two normal subgroups? – Arturo Magidin Dec 7 '20 at 3:16

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".
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 $$12$$.