List questions from simple-groups

Published on 30 Jul 2023 - 18:12

76

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The only proper normal subgroups of a nonabelian quasisimple group are subgroups of its centre.

Published on 23 Feb 2026 - 2:15

87

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Is there a way to computationally verify that the sporadic groups are simple?

Published on 25 Mar 2026 - 9:53

89

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Motivation for triple cover of $A_6$ (and $A_7$)

Published on 02 Sep 2023 - 18:06

27

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Splitting of an extension

Published on 25 Mar 2026 - 20:35

470

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Has every infinite simple group a faithful irreducible representation?

Published on 26 Mar 2026 - 19:01

47

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A reference request for $SL(2,q)$ being quasisimple for prime powers $q\ge 4$.

Published on 18 Sep 2023 - 14:09

380

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"Simple" group of order $1004913$ problem, fixed point part

Published on 26 Sep 2023 - 17:37

53

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If $G$ is simple, then $Z({\rm Aut}(G)) = \{1\} \iff G$ is non-abelian

Published on 17 Oct 2023 - 15:43

59

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Are commutator subgroups simple?

Published on 07 Nov 2023 - 17:24

112

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Diagonal in the power of a group

Published on 20 Nov 2023 - 16:51

137

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$[G:H]=4$ where $H\neq 1$ means that $G$ is not simple.

Published on 01 Dec 2023 - 3:56

107

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$G$ is a finite, simple, nonabelian group. If $|G:H| = p$, then $|\{gHg^{-1}\}|=p$

Published on 05 Dec 2023 - 16:31

98

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Elements of $\mathbb{Z}/p\mathbb{Z}$ and prove it's simple group

Published on 26 Dec 2023 - 14:52

107

Views

No simple group of order $p^nq^m$, with barely invoking Sylow theorems

Published on 25 Jan 2024 - 16:39

List Question

Does every perfect finite group have a fixed point free representation?

The only proper normal subgroups of a nonabelian quasisimple group are subgroups of its centre.

Is there a way to computationally verify that the sporadic groups are simple?

Motivation for triple cover of $A_6$ (and $A_7$)

Splitting of an extension

Has every infinite simple group a faithful irreducible representation?

A reference request for $SL(2,q)$ being quasisimple for prime powers $q\ge 4$.

"Simple" group of order $1004913$ problem, fixed point part

If $G$ is simple, then $Z({\rm Aut}(G)) = \{1\} \iff G$ is non-abelian

Are commutator subgroups simple?

Diagonal in the power of a group

$[G:H]=4$ where $H\neq 1$ means that $G$ is not simple.

$G$ is a finite, simple, nonabelian group. If $|G:H| = p$, then $|\{gHg^{-1}\}|=p$

Elements of $\mathbb{Z}/p\mathbb{Z}$ and prove it's simple group

No simple group of order $p^nq^m$, with barely invoking Sylow theorems

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