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.

  • $\begingroup$ 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. $\endgroup$ Mar 29, 2023 at 0:57
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    $\begingroup$ 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. $\endgroup$ Mar 29, 2023 at 2:19
  • $\begingroup$ 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. $\endgroup$ Mar 29, 2023 at 2:37
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    $\begingroup$ @GerryMyerson These are all special cases of what I'm mentioning though (not that there aren't any others). $\endgroup$ Mar 29, 2023 at 4:04
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    $\begingroup$ 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 $\endgroup$ Mar 30, 2023 at 1:43

2 Answers 2


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.

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    $\begingroup$ The center cannot have prime index in any group. $\endgroup$
    – YCor
    Mar 28, 2023 at 22:56
  • $\begingroup$ @Ycor sorry I missed that $N$ has index $p$. $\endgroup$ Mar 28, 2023 at 23:43
  • $\begingroup$ But I assume these references prove something, in a more flexible context, but what exactly? $\endgroup$
    – YCor
    Mar 29, 2023 at 6:07
  • $\begingroup$ @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). $\endgroup$ Mar 29, 2023 at 9:06
  • $\begingroup$ @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. $\endgroup$ Mar 29, 2023 at 9:09

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