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I have been solving some exercise in group theory and came up with two observations, the first and second below. Then I tried to generalize the fact from that exercise and got an idea of other two observations (the third and fourth below), but was unable to prove or disprove them. Any further observations about them is welcome, and a proof or disproof would resolve my problem.

Let $G$ be a finite group, let $H\leq G$ be its subgroup, and let's define the notation $[x]_A$ to be the conjugacy class of element $x\in A$ of group $A\in \{G,H\}$.

  1. It holds that $(\forall x\in H)\Big(\frac{|[x]_G|}{|[x]_G\cap H|}\text{ divides }[G:H]\Big)$.
  2. It holds that $(\forall x\in H)[x]_H\subseteq [x]_G$.
  3. Does it hold that $(\forall x,y\in H)\big(\text{some of the numbers }\frac{|[x]_G|}{|[x]_G\cap H|}\text{ and }\frac{|[y]_G|}{|[y]_G\cap H|}\text{ divides the other one}\big)$?
  4. Does it hold that $(\forall x\in H)(\forall a,b \in[x]_G\cap H)|[a]_H|=|[b]_H|$?

The exercise that I was solving was the next one:

Let $n\in\mathbb{N}$ be a natural number. And let $H\leq S_n$, $[S_n:H]<n$ be a subgroup of symmetric group $S_n$ of order less than $n$. If $n=5$, prove that $H$ is either $A_n$ or $S_n$.

To prove that, I used the first statement from above. Then I tried to generalize this exercise and replace the condition $n=5$ with $n\geq 5$ and came up with the third and fourth statements as possibly useful lemmas.

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  • $\begingroup$ Where are these exercises from? $\endgroup$
    – Shaun
    Jan 8 at 21:30
  • $\begingroup$ @Shaun, it is from one of tests from the faculty where I study $\endgroup$
    – donaastor
    Jan 8 at 21:43
  • $\begingroup$ What does "its subgroup" mean? How does $H$ belong to $x$? $\endgroup$ Jan 8 at 21:46
  • $\begingroup$ @ArturoMagidin Excuse me, I don't see what you mean. Can you tell me where is that part that confused you? $\endgroup$
    – donaastor
    Jan 8 at 21:57
  • $\begingroup$ I think it's been edited out, since there was a typo before but I don't see it now. $\endgroup$ Jan 8 at 21:58
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3 is false. Take $G=S_6$, and let $H$ be the stabilizer of $6$, which is isomorphic to $S_5$ and has index $6$ in $G$. Let $x=(1234)$; the number of elements in the conjugacy class of $x$ in $G$ is the number of $4$-cycles in $G$, which is $90$. Of those, the ones that are in $H$ are precisely the $4$-cycles that fix $6$, which is the same number as the number of $4$-cycles in $S_5$, namely $30$ of them. So your first quotient is $$\frac{|[x_G]|}{|[x_G]\cap H|} = \frac{90}{30} = 3.$$ Now let $y=(123)$. Its conjugacy class in $G$ consists of all $3$-cycles, and there are $40$ of them; the ones in $H$ are the ones that fix $6$, and there are as many of them as there are $3$-cycles in $S_5$, namely $20$ of them. So the second quotient is $$\frac{|[y_G]|}{|[y_G]\cap H|} = \frac{40}{20} = 2.$$ Of course, neither of $2$ and $3$ divide the other.

(I think the answer for 4 is "yes", but I'm still checking) As has been shown, the answer for 4 is "no" in general.

But it is true if $H$ is normal, because in that situation the $G$-conjugacy classes of elements of $H$ split into equal-sized conjugacy classes in $H$.

To see this, assume that $a,b\in H$ are conjugate in $G$, so that there exists $g\in G$ with $gag^{-1}=b$. The number of elements in $[a]_H$ is the index of its centralizer in $H$, $$C_H(a) = \{h\in H\mid ha=ah\},$$ and the number of elements in $[b]_H$ is likewise the index in $H$ of $$C_H(b)=\{h\in H\mid hb=bh\}.$$ Now consider the action of $g$ on $H$ by conjugation: I claim that $gC_H(a)g^{-1} = C_H(b)$. Indeed, if $h\in C_H(a)$, then $ghg^{-1}\in C_H(b)$: $$(ghg^{-1})b = ghg^{-1}gag^{-1} = ghag^{-1} = gahg^{-1} = gag^{-1}(ghg^{-1}) = b(ghg^{-1}).$$ Thus, $gC_H(a)g^{-1}\leq C_H(b)$. Symmetrically, $g^{-1}C_H(b)g\leq C_H(a)$, which gives the converse inclusion. Thus, $[H:C_H(a)]=[H:C_H(b)]$, and so $|[a]_H|=|[b]_H|$, as desired.

(More generally, if $\varphi$ is an automorphism of $G$, then $\varphi(C_G(a)) = C_G(\varphi(a))$, and here $\varphi$ is the automorphism of $H$ induced by conjugation by $g$.)

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  • $\begingroup$ Thank you for this solution! $\endgroup$
    – donaastor
    Jan 8 at 22:36
  • $\begingroup$ @donaastor: Note the addition: 4 becomes true if $H$ is normal. $\endgroup$ Jan 8 at 22:37
  • $\begingroup$ Yes, I went through it right now. Thank you for that observation, it really gave me some insight! $\endgroup$
    – donaastor
    Jan 8 at 22:54
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4 is false. A counterexample is given with $G=\operatorname{PGL}(3)$ and with $H≅\operatorname{GL}(2)$ the subgroup of matrices of the form $$\begin{pmatrix}1&0&0\\0&a&b\\0&c&d\end{pmatrix}\text{.}$$

Choose $x$ different from $0$ and $1$. Take $$a=\begin{pmatrix}1&0&0\\0&x&0\\0&0&x\end{pmatrix}\text{.}$$ Then $a$ is in the center of $H$ so that $[a]_H = \{a\}$. But a conjugate in $G$ is $$b = \begin{pmatrix}x&0&0\\0&x&0\\0&0&1\end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1/x\end{pmatrix}$$ which is not in the center of $H$, so that $|[b]_H| > 1$.

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  • $\begingroup$ Although I am not familiar with those grouops, thank you for confirming that 4 is false. I knew that key to resolving such questions is knowing as many examples of groups as possible. You certainly approved my expectation! $\endgroup$
    – donaastor
    Jan 8 at 22:42
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    $\begingroup$ Don't you need $x^2=1$ (and $ad-bc=1$, of course). I think this would work over $\mathbb{F}_3$ with $x=2$, of course. $\endgroup$ Jan 8 at 22:52
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    $\begingroup$ @ArturoMagidin Ah, I was thinking about the projective general group and I wrote PSL instead of PGL! Otherwise the end wouldn't work even if $x^2=1$. Thanks. $\endgroup$
    – Idéophage
    Jan 8 at 23:04
  • $\begingroup$ @donaastor You will certainly learn about these groups as you progress if you are interested. They come from linear algebra, which can be done on finite fields and thus gives finite groups. I don't really know what you mean by "approved your expectation" though. $\endgroup$
    – Idéophage
    Jan 8 at 23:06
  • $\begingroup$ "Confirmed" his expectation, perhaps. $\endgroup$ Jan 8 at 23:08

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