Question

If all Sylow subgroups of $G$ intersect in Sylow subgroups of solvable subgroup $H$, then $H$ is subnormal

score 106 · Answer 1 · 2026-03-25 17:31:12

106

Views

If all Sylow subgroups of $G$ intersect in Sylow subgroups of solvable subgroup $H$, then $H$ is subnormal

Published on 25 Mar 2026 - 17:31

score 83 · Answer 2 · 2015-11-03 22:29:29

83

Views

Reasoning about subgroups $H \le PSL(2,q)$ based on knowledge about subgroups of normalizers in it

Published on 03 Nov 2015 - 22:29

score 702 · Answer 3 · 2015-11-09 04:59:40

702

Views

Showing $\left\lvert G\right\rvert=105$ is not simple. Proof verification

Published on 09 Nov 2015 - 4:59

score 279 · Answer 4 · 2015-11-11 02:38:06

279

Views

Showing $|G|=90$ means $G$ is not simple by centraliser argument

Published on 11 Nov 2015 - 2:38

score 1k · Answer 5 · 2026-03-30 08:17:04

1k

Views

If $|G| = 120$ then $G$ has a subgroup of index $3$ or $5$ (or both)

Published on 30 Mar 2026 - 8:17

score 1k · Answer 6 · 2026-03-26 21:32:22

1k

Views

Normality of Sylow $p$-subgroup if all maximal subgroups have prime index

Published on 26 Mar 2026 - 21:32

score 320 · Answer 7 · 2015-11-17 07:59:37

320

Views

On $p$-subgroup of the general linear group over a vector space over $\mathbb{Z}_p$

Published on 17 Nov 2015 - 7:59

score 5.4k · Answer 8 · 2015-11-18 02:54:45

5.4k

Views

Group of Order 105 Having Normal Sylow 5/7 Subgroups

Published on 18 Nov 2015 - 2:54

score 7k · Answer 9 · 2015-11-19 19:05:53

7k

Views

There is no simple group of order 56

Published on 19 Nov 2015 - 19:05

score 2.8k · Answer 10 · 2015-11-21 02:21:56

2.8k

Views

Number of Non-Abelian Groups of order 21

Published on 21 Nov 2015 - 2:21

score 167 · Answer 11 · 2015-11-24 17:48:24

167

Views

Order of the normalizer of Sylow-$p$ subgroup of $A_{p+1}$.

Published on 24 Nov 2015 - 17:48

score 312 · Answer 12 · 2015-11-25 17:50:23

312

Views

Center of normalizer of cyclic group

Published on 25 Nov 2015 - 17:50

score 173 · Answer 13 · 2015-11-25 20:07:28

173

Views

$\pi$-complement of direct product of groups

Published on 25 Nov 2015 - 20:07

score 705 · Answer 14 · 2015-12-01 05:09:42

705

Views

If $P \in Syl_p(G)$ and $P$ is cyclic, then $N_G(P)=C_G(P)$

Published on 01 Dec 2015 - 5:09

score 177 · Answer 15 · 2015-12-01 20:20:47

177

Views

If $C_G(x) \leq H$ for every $p$-element $x \in H$, then $p$ cannot divide both $|H|$ and $|G:H|$

Published on 01 Dec 2015 - 20:20

List Question

If all Sylow subgroups of $G$ intersect in Sylow subgroups of solvable subgroup $H$, then $H$ is subnormal

Reasoning about subgroups $H \le PSL(2,q)$ based on knowledge about subgroups of normalizers in it

Showing $\left\lvert G\right\rvert=105$ is not simple. Proof verification

Showing $|G|=90$ means $G$ is not simple by centraliser argument

If $|G| = 120$ then $G$ has a subgroup of index $3$ or $5$ (or both)

Normality of Sylow $p$-subgroup if all maximal subgroups have prime index

On $p$-subgroup of the general linear group over a vector space over $\mathbb{Z}_p$

Group of Order 105 Having Normal Sylow 5/7 Subgroups

There is no simple group of order 56

Number of Non-Abelian Groups of order 21

Order of the normalizer of Sylow-$p$ subgroup of $A_{p+1}$.

Center of normalizer of cyclic group

$\pi$-complement of direct product of groups

If $P \in Syl_p(G)$ and $P$ is cyclic, then $N_G(P)=C_G(P)$

If $C_G(x) \leq H$ for every $p$-element $x \in H$, then $p$ cannot divide both $|H|$ and $|G:H|$

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