# A finite group whose conjugacy classes outside a normal subgroup have equal size

I would like to understand the possible structure of finite groups $$G$$ that has a normal subgroup $$N$$ of index $$p$$ (a prime) such that conjugacy classes of $$G$$ outside $$N$$ have equal size. Another way to put it is that centralizers $$C_G(g)$$ with $$g\in G-N$$ have equal order.

An obvious example: $$N$$ is abelian and $$G=N\times C_p$$ -- the direct product of $$N$$ and the cyclic group of order $$p$$. Another example is when $$G-N$$ is a single class of $$G$$. I am not aware of any other example. Any help is very much appreciated.

• Any case with $N$ abelian of prime index is always an example, since the centralizer order of $x$ and of $nx$ ($x\notin N, n \in N$) is the same. Mar 29, 2023 at 0:57
• The group of symmetries of a square has several normal subgroups of index two; for each such normal subgroup $N$, the conjugacy classes outside of $N$ have size two. Mar 29, 2023 at 2:19
• The alternating group on four letters has a normal subgroup of index three, and the two conjugacy classes outside this subgroup both have size four. The nonabelian group of order $21$ has a normal subgroup of index three, and the two conjugacy classes outside this subgroup have the same size. Mar 29, 2023 at 2:37
• @GerryMyerson These are all special cases of what I'm mentioning though (not that there aren't any others). Mar 29, 2023 at 4:04
• Here's an example that isn't covered (I think) by anything posted here yet. The special linear group of $2\times2$ matrices over the field of three elements is a group of order $24$ with a (nonabelian) normal subgroup $N$ isomorphic to the quaternion group of order eight and index three; there are four conjugacy classes outside $N$, each of size four. See people.maths.bris.ac.uk/~matyd/GroupNames/1/SL(2,3).html Mar 30, 2023 at 1:43

Here is a construction. We begin with a lemma:

Lemma. Let $$N\triangleleft G$$ be finite groups with $$p:=[G:N]$$ a prime number. Conjugacy classes of elements of $$G-N$$ have the same cardinality iff $$|N\cap C_G(g)|$$ does not change as $$g$$ varies in $$G-N$$.

Proof) For any $$g\in G-N$$, $$NC_G(g)$$ is a subgroup strictly containing $$N$$, hence $$NC_G(g)=G$$. Thus
$$C_p\cong\frac{G}{N}=\frac{NC_G(g)}{N}\cong \frac{C_G(g)}{N\cap C_G(g)}\Rightarrow |N\cap C_G(g)|=\frac{|C_G(g)|}{p}.$$

Next, let $$N$$ be a finite group and $$\phi:N\rightarrow N$$ an automorphism satisfying the following:

• $$\phi$$ is of order $$p$$, and for any $$n'\in N$$ and $$i\in\{1,\dots,p-1\}$$ one has $$|{\rm{Fix}}(u_{n'}\circ\phi^{\circ i})|=|{\rm{Fix}}(\phi)|$$ where $$u_{n'}$$ is the inner automorphism $$n\mapsto n'nn'^{-1}$$ of $$N$$ and $${\rm{Fix}}$$ denotes the set of fixed points. $$(\star)$$

(Notice that $$\phi$$ must be outer.)

Claim. When $$N$$ satisfies $$(\star)$$, the semidirect product $$G:=N\rtimes_\phi C_p$$ has the desired property.

Proof) Identify $$N$$ with an index $$p$$ subgroup of $$G$$, and let $$g_0\in G-N$$ be an element for which the induced automorphism $$n\mapsto g_0ng_0^{-1}$$ of $$N$$ coincides with $$\phi$$. Elements of $$G-N$$ may be uniquely written as $$n'g_0^i$$ where $$i\in\{1,\dots,p-1\}$$ and $$n'\in N$$. By the lemma, it suffices to show that the size of $$\{n\in N\mid n(n'g_0^i)=(n'g_0^i)n\}$$ is independent of the element chosen from $$G-N$$. Notice that $$n(n'g_0^i)=(n'g_0^i)n$$ may be written as $$n'(g_0^ing_0^{-i})n'^{-1}=n$$, or $$u_{n'}\circ\phi^{\circ i}(n)=n$$. Therefore, the preceding set is always of size $$|{\rm{Fix}}(u_{n'}\circ\phi^{\circ i})|=|{\rm{Fix}}(\phi)|$$.

We now discuss examples of $$(N,\phi)$$ which satisfy $$(\star)$$. A simple case, mentioned in the comments, is when $$N$$ is abelian. In that case, inner automorphisms are trivial and one needs to have $$|{\rm{Fix}}(\phi^{\circ i})|=|{\rm{Fix}}(\phi)| \,\forall i\in\{1,\dots,p-1\}$$ which clearly holds for any automorphism $$\phi:N\rightarrow N$$ whose order is a prime number $$p$$.

Here is a more complicated example with $$N$$ non-abelian: Suppose $$p$$ is odd, and take $$N$$ to be the group $$\Bbb{Z}/p^2\Bbb{Z}\rtimes\Bbb{Z}/p\Bbb{Z}$$ of order $$p^3$$ whose product rule is given by

$$(u,v)*(s,t)=(u+(pv+1)s,v+t).$$ We claim that the automorphism $$\phi$$ defined by $$(1,0)\mapsto (1,1)$$ and $$(0,1)\mapsto (0,1)$$ satisfies $$(\star)$$. First, notice that $$\phi$$ is of order $$p$$: A straightforward induction shows that $$\phi^{\circ k}(1,0)=(1,k),\quad \phi^{\circ k}(0,1)=(0,1).$$ In particular, $$\phi^{\circ p}$$ is identity since it fixes both $$(1,0)$$ and $$(0,1)$$. Finally, we verify $$|{\rm{Fix}}(u_{n'}\circ\phi^{\circ i})|=|{\rm{Fix}}(\phi)|$$ for $$i\in\{1,\dots,p-1\}$$. Inner automorphisms of $$\Bbb{Z}/p^2\Bbb{Z}\rtimes\Bbb{Z}/p\Bbb{Z}$$ do not change the second component while the second component of $$\phi(u,v)$$ is $$u+v$$. Comparing second components, $$u_{n'}\circ\phi^{\circ i}(u,v)=(u,v)$$ requires $$iu+v=v$$, or $$u=0$$. Conversely, any element of $$\Bbb{Z}/p\Bbb{Z}$$ is fixed by $$\phi$$ and inner automorphisms. We conclude that $${\rm{Fix}}(u_{n'}\circ\phi^{\circ i})$$ always coincide with the subgroup $$\Bbb{Z}/p\Bbb{Z}$$ of $$N=\Bbb{Z}/p^2\Bbb{Z}\rtimes\Bbb{Z}/p\Bbb{Z}$$.

The case where $$N$$ is the centre was answered in N. Ito, On finite groups with given conjugate types I, Nagoya Math. J. 6 (1953) 17–28. MR0061597 and in Ishikawa, Kenta (J-CHIBES) On finite p-groups which have only two conjugacy lengths. (English summary) Israel J. Math. 129 (2002), 119–123. I would also look at Mann, Avinoam Spreads and nilpotence class in nilpotent groups and Lie algebras. J. Algebra 421 (2015), 12–15 for generalizations.

• The center cannot have prime index in any group.
– YCor
Mar 28, 2023 at 22:56
• @Ycor sorry I missed that $N$ has index $p$. Mar 28, 2023 at 23:43
• But I assume these references prove something, in a more flexible context, but what exactly?
– YCor
Mar 29, 2023 at 6:07
• @YCor I think the first one is still relevant as there is no assumption that $G$ is a $p$-group (as far as I remember), I think it implies $G$ is nilpotent (possibly a bit more, it was many years since I have looked at it). Mar 29, 2023 at 9:06
• @YCor the second one shows that $G$ is of class at most $3$ (that was a surprise as people expected a bound only on the derived length). It might give some intuition about what result to hope for or possibly even a method. Mar 29, 2023 at 9:09