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So let $S_N$ be the symmetric group of degree $N$. We think of it as a permutation group via its natural action on the set $T=\{1,2,\ldots,N\}$.

Say that $H\leq S_N$ is a subgroup which acts transitively on $T$. However, I DONT'T WANT to assume necessarily that $H$ is primitive (that is the whole point of my question). Assume furthermore that there is an onto group homomorphism $$f:H\rightarrow S_n$$ where $n=\lfloor{N/2}\rfloor$. In fact, as was pointed out by Schmidt, the existence of this onto group homomorphism implies that $H$ is imprimitive.

In general, one cannot rule out the existence of such an $H$. For example one could have $H=S_n\ltimes\mathbf{F}_2^n$ when $N$ is even and $n=\frac{N}{2}$. Here, $S_n$ acts in the natural way by permutation on the coordinates of $\mathbf{F}_2^n$. Note that by construction, $H$ acts transitively on $T$ and it admits an onto group homomorphism on $S_n$.

Furthermore, suppose that I can produce " a lot of elements " in $H$ which contain a cycle of length $r$ in their cycle presentations (their writing as a product of disjoint cycles of $T$) for $r>n$. Then may I conclude that such an $H$ does not exist?

Q1: Is there some kind of results that would allow me to conclude that $H\supseteq A_N$, so that this would contradict the imprimitivity and therefore rule out the existence of such an $H$?

For example here is one key result which is good to know: if $H$ is assumed to be primitive and contains a cycle of length $\ell$ with $2\leq \ell\leq N-7$ ($\ell$ not necessarily prime) then combining classical results on permutation group theory one may show that $H\supseteq A_N$. However, since I assume that in my setting $H$ is imprimitive I cannot apply this result.

Q2: Do we have a good understanding of the tree of subgroups of $S_N$, especially the maximal subgroups?

Q3: Is there some kind of probabilistic result that could be used in my context?

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So let $S_N$ be the symmetric group of degree $N$. We think of it as a permutation group via its natural action on the set $T=\{1,2,\ldots,N\}$.

Say that $H\leq S_N$ is a subgroup which acts transitively on $T$. However, I DONT'T WANT to assume necessarily that $H$ is primitive (that is the whole point of my question). Assume furthermore that there is an onto group homomorphism $$f:H\rightarrow S_n$$ where $n=\lfloor{N/2}\rfloor$. In fact, as was pointed out, the existence of this onto group homomorphism implies that $H$ is imprimitive.

In general, under these two conditions one cannot conclude that $H=S_N$ (or that rule out the existence of such an $H\supseteq A_N$), for H$. For example , one could have$H=S_n\ltimes\mathbf{F}_2^n$when$N$is even and$n=\frac{N}{2}$. Here,$S_n$acts in the natural way by permutation on the coordinates of$\mathbf{F}_2^n$. Note that by construction,$H$acts transitively on$T$and it admits an onto group homomorphism on$S_n$. Furthermore, suppose that I can produce " a lot of elements " in$H$which contain a cycle of length$r$in their cycle presentations (their writing as a product of disjoint cycles of$T$) for$r>n$. Then may I conclude that such an$H$does not exist? Q1: Then is Is there some kind of results that would allow me to conclude that$H\supseteq A_N$?, so that this would contradict the imprimitivity and therefore rule out the existence of such an$H$? For example , here is one key result which is good to know: if$H$is assumed to be primitive and contains a cycle of length$\ell$with$2\leq \ell\leq N-7$($\ell$not necessarily prime) then combining classical results on permutation group theory one may show that$H\supseteq A_N$. Unfortunately However, since I don't know assume that$H$is primitive!imprimitive I cannot apply this result. Q2: Do we have a good understanding of the tree of subgroups of$S_N$, especially the maximal subgroups? Q3: Is there some kind of probabilistic result that could be used in my context? 3 edited body So let$S_N$be the symmetric group of degree$N$. We think of it as a permutation group via its natural action on the set$T=\{1,2,\ldots,N\}$. Say that$H\leq S_N$is a subgroup which acts transitively on$T$. However, I DONT'T WANT to assume necessarily that$H$is primitive (that is the whole point of my question). Assume furthermore that there is an onto group homomorphism $$f:H\rightarrow S_n$$ where$n=\lfloor{N/2}\rfloor$. In general, under these two conditions one cannot conclude that$H=S_N$(or that$H\supseteq A_N$), for example, one could have$H=S_n\rtimes\mathbf{F}_2^n$H=S_n\ltimes\mathbf{F}_2^n$ when $N$ is even and $n=\frac{N}{2}$. Here, $S_n$ acts in the natural way by permutation on the coordinates of $\mathbf{F}_2^n$. Note that by construction, $H$ acts transitively on $T$ and it admits an onto group homomorphism on $S_n$.

Furthermore, suppose that I can produce " a lot of elements " in $H$ which contain a cycle of length $r$ in their cycle presentations (their writing as a product of disjoint cycles of $T$) for $r>n$.

Q1: Then is there some kind of results that would allow me to conclude that $H\supseteq A_N$?

1. For example, if $H$ is primitive and contains a cycle of length $\ell$ with $2\leq \ell\leq N-7$ ($\ell$ not necessarily prime) then combining classical results on permutation group theory one may show that $H\supseteq A_N$. Unfortunately I don't know that $H$ is primitive!

Q2: Do we have a good understanding of the tree of subgroups of $S_N$, especially the maximal subgroups?

Q3: Is there some kind of probabilistic result that could be used in my context?

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