# Let $G$ be a group with $n$ conjugacy classes. Prove there is a maximum $2^{n-1}$ normal subgroups.

Let $$G$$ be a group with $$n$$ conjugacy classes.

Prove there is a maximum $$2^{n-1}$$ normal subgroups.

I know the cosets of normal subgroup $$H$$ with index $$j$$ has the form: $$g_1H,g_2H,\cdots ,g_jH , g_i\in G$$

In addition , I think if $$\exists g\in G ,g^{-1}H_1g=H_2$$, such that $$H_1,H_2$$ are normal subgroups so they have the same conjugacy classes.

I don't know how to approach the problem.

Any help is welcome,Thanks !

Hint: every normal subgroup is the disjoint union of some of the conjugacy classes and the conjugacy class $$\{1\}$$ is always part of that.

• There are $n-1$ conjugacy classes except $\{1\}$. Suppose $N$ is a normal subgroup , $1$= the conjugacy class is part of the disjoint union of some of the conjugacy classes of $N$ , $0$=the conjugacy class is *not* part of the disjoint union of some of the conjugacy , classes of $N$ , there are 2^{n-1} different options , hence there are maximum $2^{n-1}$ ? Is it correct ?
– algo
Jun 18 at 20:32
• Yes you got it, well done! Jun 18 at 21:01

If $$n = 1$$, then $$G = \{e\}$$ because no non-identity element is every conjugate to the identity element. Thus, in this case, we see equality happens. That is, the number of normal subgroups is equal to $$2^{n-1}$$.

Likewise, if $$n=2$$, then equality happens. This is because if $$n=2$$, the group $$G$$ is non-trivial, and so it has $$2 = 2^{2-1}$$ normal subgroups: itself and the identity subgroup.

I claim that these are the only cases where equality is achieved.

Suppose $$G$$ is a group with $$n$$ conjugacy classes with $$n\geq 3$$. Then it has fewer than $$2^{n-1}$$ normal subgroups.

Proof: We prove it by contradiction. So, assume $$G$$ is a group with $$n\geq 3$$ conjugacy classes and $$2^{n-1}$$ normal subgroups. Since $$n\geq 3$$, we can find three distinct conjugacy classes. Call them $$A_0 =\{e\}$$, $$A_1$$, and $$A_2$$.

From Nicky's answer, each of $$H_1 = A_0\cup A_1$$, $$H_2 = A_0\cup A_2$$, and $$H_{12}=A_0\cup A_1\cup A_2$$ must be a normal subgroup of $$G$$. Choose $$g_1\in A_1$$ and $$g_2\in A_2$$.

Since $$H_1$$ is a subgroup of $$G$$ and $$g_1\in H_1$$, $$g_1^{-1}\in H_1$$. But note that $$g_1\neq e$$, so $$g_1^{-1}\neq e$$, so $$g_1^{-1} \in A_1$$. Analogously, $$g_2^{-1}\in A_2$$.

Since $$H_{12}$$ is a subgroup of $$G$$, $$g_1 g_2\in H_{12}$$. Notice that $$g_1g_2\notin A_1$$, for if it was, then $$g_2 = g_1^{-1}(g_1g_2)\in A_1$$. Likewise, $$g_1g_2\notin A_2$$. Thus, we conclude $$g_1g_2\in A_0 = \{e\}$$. In other words, $$g_2 = g_1^{-1}$$. Thus $$g_2\in A_2 \cap A_1$$, giving a contradiction. $$\square$$

So, if $$n\geq 3$$ is the number of conjugacy classes of a group $$G$$, then the number of normal subgroups is at most $$2^{n-1}-1$$. This bound can actually be achieved: consider $$G = D_6$$, the dihedral group of order $$6$$. This has $$n=3$$ conjugacy classes (the identity, the two non-trivial rotations, and all the reflections), and it has $$2^{3-1} - 1 = 3$$ normal subgroups: the trivial subgroup, the subgroup of all rotations, and the whole group.

On the other hand, not all groups with $$n=3$$ have $$3$$ normal subgroups: consider $$\mathbb{Z}/3\mathbb{Z}$$.