4
$\begingroup$

Let $G$ be a finite group, $G',H$ be its subgroups and $H'=G'\cap H$. For each $g\in G$, we create a map $f_g:G'/H'\rightarrow G/H: aH'\rightarrow gaH$. It's easy to see that the map is well defined and injective. Let $S$ be a subset of $G/H$, assume that there is no $g\in G$ such that $f_g(G'/H')\subset S$.

Question: I want to estimate and find some properties of $M(G,G',H)=\max|S|$ and $\alpha(G,G',H)=\max\frac{|G/H|}{|G/H|-|S|}$, for all or some particular cases of groups $G,G',H$. Note that we have $M(G,G',H)=\left(1-\frac{1}{\alpha(G,G',H)}\right)|G/H|$.

Some results that I have found:

a) $1\leq\alpha(G,G',H)\leq |G'/H'|$. The first inequality is trivial. Assume that $|S|> \left(1-\frac{1}{|G'/H'|}\right)|G/H|$ then:

$\mathbb{E}_{g\in G}[|f_g(G'/H')\cap S|]=|G'/H'|\mathbb{P}_{a\in G'/H',g\in G}[f_g(a)\in S]=\frac{|G'/H'|}{|G/H|}|S|>|G'/H'|-1 $

So there exists $g\in G$ such that $|f_g(G'/H')\cap S|=|G'/H'|\Rightarrow f_g(G'/H')\subset S$.

b) If $G'\subset H\Rightarrow H'=G'$, then by a), $\alpha(G,G',H)=1,M(G,G',H)=0$.

c) If $H,H'$ is a trivial group then $\alpha(G,G',H)=|G'|$ because each coset of $G/G'$ contains at most $|G'|-1$ elements of $S$, so $|S|\leq (|G'|-1)|G/G'|$

d) If $N$ is a normal subgroup of $G,H$ and $N'=N\cap G'$ then $\alpha(G/N,G'/N',H/N)=\alpha(G,G',H),M(G/N,G'/N',H/N)=M(G,G',H)$, because the two set of left cosets and the set of all function $f_g$ are still the same under isomorphism after taking quotent by $N$ and $N'$.

e) By c), d), if $G'$ is a commutative group then $\alpha(G,G',H)=\alpha(G/H,G'/H',{e})=|G'/H'|$.

f) If $N$ is a subgroup of $G,G'$ such that $N\cap H$ be the trivial group and $nh=hn,\forall n\in N, h\in H$ then $\alpha(G,G',H)=|N|\alpha(G,G',NH)$ (I proof by dividing $G/NH,G'/NH'$ into $|N|$ parts but the proof is quite complicative so let's omit it).

g) If $H'$ is a trivial group and $hg=gh, \forall h\in H, g\in G'$ then $\alpha(G,G',H)=|G'|\alpha(G,G',G'H)=|G'|$ by f), b).

h) $M(G,G',H)=|G/G'H|M(G'H,G',H), \alpha(G,G',H)=\alpha(G'H,G',H)$

so we can assume $G=G'H$.

Motivation: Let $P_S$ be a group of permutation of the set $S$. Let $\{1,2,...,n\},n\geq 3$ be the sets of vertices of the complete graph $K_n$, we have a bijection from the set of left cosets $P_{\{1,2,...,n\}}/(P_{\{3,4,...,n\}}\times P_{\{1,2\}})$ to the set of edges of $K_n$ by map the coset $\sigma(P_{\{3,4,...,n\}}\times P_{\{1,2\}})$ to the edge $(\sigma(1),\sigma(2))$.

For each $\sigma\in P_{\{1,2,...,n\}}$, we see that the map $f_{\sigma}: P_{\{1,2,...,r\}}/(P_{\{3,4,...,r\}}\times P_{\{1,2\}})\rightarrow P_{\{1,2,...,n\}}/(P_{\{3,4,...,n\}}\times P_{\{1,2\}})$ corresponds to the complete subgraph $K_r$ of $K_n$ with vertices $\sigma(1),\sigma(2),...\sigma(r)$.

The subset $S$ of $P_{\{1,2,...,n\}}/(P_{\{3,4,...,n\}}\times P_{\{1,2\}})$ such that there is no $\sigma\in P_{\{1,2,...,n\}}$ such that $f_{\sigma}(P_{\{1,2,...,r\}}/(P_{\{3,4,...,r\}}\times P_{\{1,2\}}))\subset S$ creates a $K_r$-free graph of $n$ vertices, so $M(P_{\{1,2,...,n\}},P_{\{1,2,...,r\}},P_{\{3,4,...,n\}}\times P_{\{1,2\}})= \left(1-\frac{1}{r}+o(1)\right)\frac{n^2}{2}\Rightarrow \alpha(P_{\{1,2,...,n\}},P_{\{1,2,...,r\}},P_{\{3,4,...,n\}}\times P_{\{1,2\}})=r+o(1)$ with $r$ fixed and $n$ increase by Turán's theorem.

More questions:

  1. When $\alpha(G,G',H)=|G'/H'|$?

  2. Improve the lower bound of $\alpha(G,G',H)$ which is only depend on $|G'/H'|$ or show such bound doesn't exist.  

We can view the bipartite graph $K_{\{1,2,...,r\},\{r+1,r+2,...,2r\}}$ as $Aut(K_{\{1,2,...,r\},\{r+1,r+2,...,2r\}})/Aut(K_{\{1,2,...,r-1\},\{r,r+1,...,2r-2\}})$, then by Erdős–Stone-Simonovits theorem we have

$M(P_{\{1,2,...,n\}},Aut(K_{\{1,2,...,r\},\{r+1,r+2,...,2r\}}),P_{\{3,4,...,n\}}\times P_{\{1,2\}})=o(n^2)\Rightarrow \alpha(P_{\{1,2,...,n\}},Aut(K_{\{1,2,...,r\},\{r+1,r+2,...,2r\}}),P_{\{3,4,...,n\}}\times P_{\{1,2\}})=1+o(1)$

as $r$ fixed and $n$ increase, so such bound in question 2 doesn't exist and we have a new question:

2'. Is it true that for all group $G'$, subgroup $H'$ of $G'$ and $\varepsilon>0$, there exists group $G$ and subgroup $H$ of $G$ such that $M(G,G',H)$ is defined and $M(G,G',H)<\varepsilon|G/H|$?

$\endgroup$
3
  • 1
    $\begingroup$ It seems to me that the formula from the third line should be $f_g(G'/H')\subset S$. $\endgroup$
    – kabenyuk
    Dec 13, 2022 at 4:04
  • 1
    $\begingroup$ Thank you very much @kabenyuk $\endgroup$ Dec 13, 2022 at 9:53
  • 1
    $\begingroup$ Please be aware that every edit of a question or of one of its answers bumps the thread to the front page. This has happened for this thread already more than 10 times within just a few days, and this is a nuisance for other users. Please refrain from unnecessary edits to your posts. -- Usually, the vast majority of minor edits can be avoided by writing and proofreading a question or an answer (or any update to such) carefully before posting it. $\endgroup$
    – Stefan Kohl
    Dec 14, 2022 at 22:40

1 Answer 1

0
$\begingroup$

Question 2': We choose $G=P_{G'/H'\cup A},H=P_{\{(G'/H')-\{eH'\})\cup A}$, use natural acting of $G'$ on $G'/H'$, we can view $G'$ as subgroup of $G$ and $G/H=G'/H'\cup A$. We have the stabilizer subgroup with respect to the left coset $eH'$ of $G$ and $G'$ are $H$ and $H'$ respectively so $H'=G'\cap H$. Let $S\in G/H$, if $|S|\geq |G'/H'|$, because $G$ is the permutation group of $G/H$ so there exists $g\in G$ such that $f_g(G'/H')\subset S$. So $M(G,G',H)=|G'/H'|-1$, now we just need to choose $A$ such that $\varepsilon|G/H|>|G'/H'|-1$.

Question 1:

Assume $\alpha(G,G',H)=|G'/H'|$, take $S$ statisfies the condition and $|S|=(1-\frac{1}{|G'/H'|})|G/H|$. We have:

$\mathbb{E}_{g\in G}[|f_g(G'/H')\cap S|]=\frac{|G'/H'|}{|G/H|}|S|=|G'/H'|-1$

but $|f_g(G'/H')\cap S|\leq|G'/H'|-1$ so $|f_g(G'/H')\cap S|=|G'/H'|-1,\forall g\in G$. Because $gS$ also has that properties of $S$ for $g\in G$, so we can assume $\{eH\}\notin S$.

Let $R=G/H-S$, then $eH\in R$,$|R|=\frac{|G/H|}{|G'/H'|}$ and $|f_g(G'/H')\cap R|=1, \forall g\in G$. If $\{r\}=f_g(G'/H')\cap R$ then we take $h(r)=f_{hg}(G'/H')\cap R$. We see that this is a transitive action of $G$ on $R$. We have $g(eH)=eH$ if and only if $gH\in G'/H'\Rightarrow g\in \{xy|x\in G',y\in H\}$ so $\{xy|x\in G',y\in H\}$ must be a subgroup of $G$ (stable group of $eH\in R$).

Now if $\{xy|x\in G',y\in H\}$ is a subgroup of $G$, we have $G'H=\{xy|x\in G',y\in H\}$ and $|G'H|=\frac{|G'||H|}{|H'|}$ (Example 3.25 Page 15). Take $L$ be the left transversal for $G'H$ in $G$ and take $R=\{lH|l\in L\}$, it can be check that $|R|=\frac{|G/H|}{|G'/H'|}$ and $|f_g(G'/H')\cap R|=1, \forall g\in G$, we take $S=G/H-R$ then we have $\alpha(G,G',H)=|G'/H'|$.

So $\alpha(G,G',H)=|G'/H'|$ if and only if $\{xy|x\in G',y\in H\}$ is a subgroup of $G$. This result might suggest duality property of this problem.

$\endgroup$

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.