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6 votes

Let $G$ be a non-abelian finite group whose all non-linear irreducible characters are faithful. Is there a classification of these groups?

A buzzword here is "just nonabelian". In general group is called just $X$ if it has property $X$ but all its proper quotients do not have property $X$. Therefore a group is just nonabelian ...
Sean Eberhard's user avatar
6 votes
Accepted

Character values with absolute value $1$ are roots of unity

Here is a proof and in fact more information. Theorem Let $G$ be a finite group and $\chi$ a complex character of $G$, and $g \in G$. Then the following hold. (a) $|\chi(g)|=1$ if and only if $\chi(g)...
Nicky Hekster's user avatar
1 vote

Character table for a covering group of $\mathbb{Z}_n \times \mathbb{Z}_m$

Since you gave the [gap] tag, here is how you could compute it in GAP: ...
ahulpke's user avatar
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1 vote

Kernel of a (complex) character.

Given a character $\chi$ of the finite group $G$, the kernel of $\chi$ is the kernel of any representation $f:G \to \mathrm{GL}(V)$ affording $\chi$. The sum of characters is afforded by the direct ...
Stephen's user avatar
  • 14.9k
1 vote
Accepted

What is the action on $\mathrm{Sp}_2(q^2)$ which makes $\mathrm{Sp}_2(q^2)\colon 2$ a maximal subgroup of $\mathrm{Sp}_4(q)$ for an even power of $q$?

Let $F= {\mathbb F}_q$ and $K={\mathbb F}_{q^2}$. You can embed $K \setminus \{0\}$ into ${\rm GL}(2,F)$ my mapping a primitive element $w$ of $K$ to the companion matrix $M$ of its minimal ...
Derek Holt's user avatar
1 vote

Character values with absolute value $1$ are roots of unity

Reduce to: let $\omega$ be a root of $1$ and $P(x)\in \mathbb{Z}[x]$ such that $|P(\omega)|=1$. Then $P(\omega)$ is again a root of $1$. For $\omega^n=1$ for some $n$, so $\omega$ is a root of an (...
orangeskid's user avatar
  • 54.3k

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