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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 $M(G,G',H)<\varepsilon|G/H|$?

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    $\begingroup$ It seems to me that the formula from the third line should be $f_g(G'/H')\subset S$. $\endgroup$
    – kabenyuk
    Commented Dec 13, 2022 at 4:04
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    $\begingroup$ Thank you very much @kabenyuk $\endgroup$ Commented Dec 13, 2022 at 9:53
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    $\begingroup$ 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
    Commented Dec 14, 2022 at 22:40

1 Answer 1

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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.

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