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In the case that $n$ is prime and $H$ is normal in $G$. Your second questions has an affirmative answer. The composition length is $1$, $n$, or $1+d$ where $d\mid (n-1)$, in each case at most $n$. See …

I take it Pablo your question can be rephrased as follows. Does there exist an epimorphism $\tau\colon A\ltimes C\to A$ where $A$ acts irreducibly on $C$ and where $\ker(\tau)\ne C$? If this is your q …

One uses a matrix version of Hilbert's Satz 90 to answer Geoff's comment "If the field $E$ is not this minimal field, it seems less obvious to me how to realise the representation over the subfield ge …

The answer to this question is No. Let $U$ be the natural module for $H=\textrm{SL}(2,5^5)$ and let $V=U\otimes U^\sigma\otimes\cdots\otimes U^{\sigma^4}$ where $\sigma$ is field automorphism $\lambda …

Imprecise question: To get a normalizing field outer automorphism of order $r$, must we multiply the dimension by $r$? Precise hypothesis: Let $p\geqslant 5$ be a prime, let $q$ be a power of $p$ and …

Schur's lemma has a different generalization when the coefficient field $F$ is not algebraically closed. Then you get $M_{m_1}(D_1)\times\cdots\times M_{m_k}(D_k)$ where $D_i:={\rm Hom}_{FG}(\rho_i,\r …

This answer may be more general than you need. Let $F$ be a field, and let $G=\langle g_1,\dots,g_m\rangle$ be a subgroup of ${\rm GL}(n,F)$ which preserves a form, with matrix $J$. Then $g_i J g_i^{\ …

This is an interesting question, even though it is not well defined. Call a group "good" if it has a "good" bijection between its conjugacy classes and its irreducible complex representations. I agree …

First let me paraphrase the question. Given integers $m,n,k$ each at least 2, set $d:=\max(m,n,k)+2$. Do there exist elements $a,b$ in the symmetric group $S_d$ such that $|a|=m$, $|b|=n$ and $|ab|=k$ …

The comment by Frieder Ladisch suggests to me that considering exponents may be relevant. Suppose that we generalize Stefan Kohl's function $f_p(n)$ as follows: Definition: Fix a prime $p$ and an exp …

If $n=2$ the quadratic map $\mathbb{R}^2\to\mathbb{R}^2$ with $(x_1,x_2)\mapsto (x_1^2-x_2^2,2x_1x_2)$ is surjective. This follows because the map $\mathbb{C}\to\mathbb{C}$ with $z\mapsto z^2$ is surj …

@Marty Isaacs: There exist non-nilpotent groups whose conjugacy class sizes are all squares. For example, let $G$ be Magma's 93rd group order 540. It has class sizes 1,4,9. Indeed, |G|=15*1+30*4+45*9. …

I have a question regarding a partial order $<$ on the set ${\rm Part}(n)$ of partitions of $n$. Given $\lambda=(\lambda_1,\lambda_2,\ldots)\in{\rm Part}(n)$ with $\sum_{i\geq1} \lambda_i=n$ and $\lam …