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 !


2 Answers 2


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

  • 3
    $\begingroup$ 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 ? $\endgroup$
    – algo
    Jun 18 at 20:32
  • 2
    $\begingroup$ Yes you got it, well done! $\endgroup$ 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 $\mathbb{Z}/3\mathbb{Z}$.


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