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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 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?

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 $H=S_n\ltimes\mathbf{F}_2^n$ where $N$ is even and $n=\frac{N}{2}$. We let $H$ act on $T$ in the following way: We divide $T$ in $n$ disjoint blocks of size $2$. We let $S_n$ permute the $n$ blocks without swapping the pair in each block, and we let $\mathbf{F}_2^n$ permute (resp. acts like the identity) the two elements in the i-th block if the i-th coordinate of an element $\sigma\in \mathbf{F}_2^n$ is $\overline{1}$ (resp. $\overline{0}$). It thus follows that $H$ acts transitively (but imprimitively) on $T$.

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 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?

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, 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 $H$ 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?

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 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?

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 $H=S_n\ltimes\mathbf{F}_2^n$ 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?

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, 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 $H$ 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?