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Here is a very natural question: Q: Is it always possible to generate a finite simple group with only $2$ elements? In all the examples that I can think of the answer is yes. If the answer is posit …

Let $q$ be a power of $2$ and let $(V,Q)$ be a quadratic space of dimension $2m$ over $\mathbb{F}_q$. Up to isometry, we know that we have exactly two classes of such quadratic spaces: the plus type a …

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 transiti …

Let $G$ be a nontrivial finite group. Given $n\in\mathbb{Z}_{\geq 1}$, let $G^n$ be the cartesian product of $n$ copies of $G$. Further let $S\subseteq G^n$ be a generating set of $G^n$. Question: Do …

Well I think I have more or less an answer to my question. I have shown that the set of all maximal imprimitive transitive subgroups $H\leq S_N$ is of the form $$ S_{N/r}^{r}\rtimes S_r $$ for $r|N$ …

Recently I learned that the cardinality of a minimal set of generators of a finite $p$-group $G$ is well defined namely it is equal to the dimension of $H^1(G,\mathbb{F}_p)$. Moreover, if $S:=\{g_1,\l …

The impossibility also follows from Jordan's lemma: Let $G$ be a finite group which acts transitively on a set $\Omega$ with $|\Omega|:=n\geq 2$. Then there exists a $g\in G$ such that $\chi(g)=0$ wh …