Every normal subgroup $N$ of a group $G$ is a union of conjugacy classes. Since every subgroup contains the identity, and the identity is in a class by itself, every normal subgroup already contains the conjugacy class of the identity.

So when is a normal subgroup comprised of exactly two conjugacy classes?

$N = \{1\} \cup \mathcal K$

Here is what I see so far:

  1. Unless $|N|=2$ and $N \leq Z(G)$, the subgroup must have trivial intersection with the center, since each element in the center is contained in its own conjugacy class.
  2. Since $|\mathcal K|$ is the index of the centralizer $C_G(k)$ of any $k\in\mathcal K$, and $|G:C_G(k)| = |N|-1$ divides $|G|$, we must have that G is divisible by the product $|N|(|N|-1)$ of two consecutive numbers. This also implies $|G|$ is even.

Any inner automorphism fixes $N$, but I don't know about outer automorphisms, so $N$ may not have to be a characteristic subgroup.

What is the full characterization of these types of normal subgroups? Do they have any important properties?

Edit: Ted is correct.

  • $\begingroup$ In the group $S_3$, the subgroup of order 3 is normal and consists of the union of 2 conjugacy classes. So the conclusion of 1 is not correct. $\endgroup$ – Ted Jul 28 '12 at 18:23
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    $\begingroup$ Yeah, #1 should only conclude G is not nilpotent. N is definitely a p-group. $\endgroup$ – Jack Schmidt Jul 28 '12 at 18:27
  • $\begingroup$ It must in fact have exponent p, since it would otherwise contain elements of more than 2 different orders, making it impossible to contain only two conjugacy classes. $\endgroup$ – Tobias Kildetoft Jul 28 '12 at 18:35
  • $\begingroup$ N does not have to be characteristic. Replace $G$ with $G \times G$. The two copies of $N$ (both of which consists of exactly two conjugacy classes) are exchanged by an automorphism of $G\times G$ (that just switches the factors). $\endgroup$ – Jack Schmidt Jul 28 '12 at 18:46
  • $\begingroup$ Observation 2 hints that G acts 2-transitively on its regular normal (elementary abelian) subgroup N, so groups like AGL(1,N) are natural examples. $\endgroup$ – Jack Schmidt Jul 28 '12 at 18:47

All groups are finite. The following is an old result of Wielandt, I believe.

Proposition: Such a subgroup $N$ must be an elementary abelian $p$-group, and every elementary abelian $p$-group is such a subgroup in a certain group $G$.

Proof: If two elements of $G$ are conjugate, then they have the same order. Hence every non-identity element of $N$ has the same order. The order cannot be composite since $g^a$ has order $b$ if $g$ has order $ab$. Hence $N$ is a $p$-group. The commutator subgroup of $N$ is characteristic in $N$ and so normal in $G$. Hence it is either all of $N$ or just $1$; however, in a non-identity $p$-group the commutator subgroup is always a proper subgroup. In particular, $N$ is abelian and every element has order $p$.

Now suppose such an elementary abelian group $N$ is given. Let $G=\operatorname{AGL}(1,p^n)$ be the set of affine transformations of the one-dimensional vector space $K$ over the field $K$ of $p^n$ elements. That is $G$ consists of all $\{ f : K \to K :x \mapsto \alpha x + \beta ~\mid~ \alpha,\beta \in K, \alpha \neq 0 \}$. Then every non-identity element of $N=K_+=\{ f : K \to K : x \mapsto x + \beta ~\mid ~ \beta \in K \}$ is conjugate under $K^\times = \{ f : K \to K : x \mapsto \alpha x ~\mid~ \alpha \in K, \alpha \neq 0 \}$. Thus the proof is complete. $\square$.

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  • $\begingroup$ Let me know if you want the book reference. It is in his Permutation Groups book, which is worth reading cover to cover anyways. $\endgroup$ – Jack Schmidt Jul 28 '12 at 18:35

Your point (1) cannot be correct, as follows from
$$G=A_4\,\,,\,N:=\{(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\}$$ Here, $\,N\lhd G\,$ , indeed $\,N\cap Z(G)=1\,$ (which is not big deal as the center of this alternating group is trivial), but $\,N\,$ is the Sylow $\,2-\,$ subgroup of $\,A_4\,$ .

Number 2 is, imo, correct.

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  • $\begingroup$ Yes, you are correct. I have taken a moment to edit the mistake, as well as to reflect upon my hastiness. $\endgroup$ – PrimeRibeyeDeal Jul 28 '12 at 18:32
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    $\begingroup$ math.stackexchange.com/questions/175729/… is meant to help include the more general family of AGL(1,q) into one's collection of examples $\endgroup$ – Jack Schmidt Jul 28 '12 at 18:37

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