4
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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

3
$\begingroup$

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 $N=\Bbb{Z}/p^2\Bbb{Z}\rtimes\Bbb{Z}/p\Bbb{Z}$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.