Question

Some very short clarification on quotient groups

score 66 · Answer 1 · 2015-05-09 11:23:52

66

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Some very short clarification on quotient groups

Published on 09 May 2015 - 11:23

score 155 · Answer 2 · 2015-05-09 11:56:33

155

Views

Proving Quotient group is cyclic / abelian via Isomorphism theorem

Published on 09 May 2015 - 11:56

score 14.7k · Answer 3 · 2015-05-10 11:33:03

14.7k

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Is every subgroup of a normal subgroup normal?

Published on 10 May 2015 - 11:33

score 18.4k · Answer 4 · 2015-05-16 11:18:10

18.4k

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Prove every group of order less or equal to five is abelian

Published on 16 May 2015 - 11:18

score 247 · Answer 5 · 2015-05-19 16:24:21

247

Views

If $G=S_4, N=\{e;(1,2)(3,4);(1,3)(2,4);(1,4)(2,3)\}$ show that $N\lhd G$

Published on 19 May 2015 - 16:24

score 1.3k · Answer 6 · 2026-03-31 10:07:41

1.3k

Views

The Image of normal subgroup is also normal subgroup?

Published on 31 Mar 2026 - 10:07

score 118 · Answer 7 · 2015-05-24 21:10:28

118

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prove that if $N\lhd G$, $ M\lhd G$, $M\bigcap N=\{e\}$ so: $mn=nm , \forall n \in N,\forall m\in M$

Published on 24 May 2015 - 21:10

score 98 · Answer 8 · 2015-05-26 09:44:03

98

Views

$H,N(H)$ are subgroups of $G$ show that $H\lhd N(H)$

Published on 26 May 2015 - 9:44

score 2.9k · Answer 9 · 2015-05-28 12:19:35

2.9k

Views

Is $O(n)$ normal in $GL(n)$?

Published on 28 May 2015 - 12:19

score 434 · Answer 10 · 2015-05-31 23:01:02

434

Views

Disjoint normal subgroups - one contained in the centralizer of the other

Published on 31 May 2015 - 23:01

score 4.9k · Answer 11 · 2015-06-01 03:19:32

4.9k

Views

Order and index of a normal subgroup $N$ are relatively prime

Published on 01 Jun 2015 - 3:19

score 277 · Answer 12 · 2015-06-01 07:32:42

277

Views

Show that $D$ is a normal subgroup.

Published on 01 Jun 2015 - 7:32

score 578 · Answer 13 · 2026-03-31 12:12:46

578

Views

Prove that all normal subgroup definitions are equivalent.

Published on 31 Mar 2026 - 12:12

score 1.6k · Answer 14 · 2015-06-10 15:21:18

1.6k

Views

$G$ is a group and $N,M$ are normal subgroups of $G$. Prove that $nm=mn$ for all $n\in N,m\in M$.

Published on 10 Jun 2015 - 15:21

score 299 · Answer 15 · 2015-06-11 09:47:10

299

Views

Given a group $G$ and subgroup $H<G$ and $g \in G$ such that $gHg^{-1} \subset H$ prove

Published on 11 Jun 2015 - 9:47

List Question

Some very short clarification on quotient groups

Proving Quotient group is cyclic / abelian via Isomorphism theorem

Is every subgroup of a normal subgroup normal?

Prove every group of order less or equal to five is abelian

If $G=S_4, N=\{e;(1,2)(3,4);(1,3)(2,4);(1,4)(2,3)\}$ show that $N\lhd G$

The Image of normal subgroup is also normal subgroup?

prove that if $N\lhd G$, $ M\lhd G$, $M\bigcap N=\{e\}$ so: $mn=nm , \forall n \in N,\forall m\in M$

$H,N(H)$ are subgroups of $G$ show that $H\lhd N(H)$

Is $O(n)$ normal in $GL(n)$?

Disjoint normal subgroups - one contained in the centralizer of the other

Order and index of a normal subgroup $N$ are relatively prime

Show that $D$ is a normal subgroup.

Prove that all normal subgroup definitions are equivalent.

$G$ is a group and $N,M$ are normal subgroups of $G$. Prove that $nm=mn$ for all $n\in N,m\in M$.

Given a group $G$ and subgroup $H<G$ and $g \in G$ such that $gHg^{-1} \subset H$ prove

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