Question

If$(ab)^n=a^nb^n$ & $(|G|, n(n-1))=1$ then $G$ is abelian

score 5.1k · Answer 1 · 2014-09-12 14:06:36

5.1k

Views

If$(ab)^n=a^nb^n$ & $(|G|, n(n-1))=1$ then $G$ is abelian

Published on 12 Sep 2014 - 14:06

score 89 · Answer 2 · 2014-09-13 06:27:15

89

Views

$\ker (T \otimes id_{Z})=\ker(T)\otimes Z$

Published on 13 Sep 2014 - 6:27

score 401 · Answer 3 · 2014-09-14 12:14:56

401

Views

Complement but not direct summand; Lam, Lectures on Modules and Rings, Example 6.17(5)

Published on 14 Sep 2014 - 12:14

score 41 · Answer 4 · 2014-09-14 12:59:45

41

Views

$G$ a finite group $n$-abelian goup and g.c.d.$\big(|G|,n(n-1)\big)=1$ , then to show $G$ is abelian

Published on 14 Sep 2014 - 12:59

score 126 · Answer 5 · 2014-09-16 10:59:29

126

Views

Direct limits of free abelian groups and diagonalization

Published on 16 Sep 2014 - 10:59

score 146 · Answer 6 · 2014-09-19 07:58:32

146

Views

$(\mathbb C^{\times}, \cdot)$ is a subgroup of $(GL(n,\mathbb C), \cdot)$

Published on 19 Sep 2014 - 7:58

score 664 · Answer 7 · 2014-09-20 14:50:41

664

Views

Groups of Order 2 with subgroups

Published on 20 Sep 2014 - 14:50

score 57 · Answer 8 · 2014-09-22 03:16:38

57

Views

Property of abelian groups without using Lagrange's theorem

Published on 22 Sep 2014 - 3:16

score 1.8k · Answer 9 · 2014-09-23 20:19:25

1.8k

Views

The number, up to isomorphism, of abelian groups of order 40 is

Published on 23 Sep 2014 - 20:19

score 1k · Answer 10 · 2014-09-24 23:33:35

1k

Views

Order of Group with Elements of Order 2

Published on 24 Sep 2014 - 23:33

score 3k · Answer 11 · 2014-09-25 04:24:17

3k

Views

Let G be a group and a; b ∈ G. Suppose |a| = |b| = |ab| = 2. Then show that ab = ba.

Published on 25 Sep 2014 - 4:24

score 260 · Answer 12 · 2014-09-25 05:10:03

260

Views

Suppose G is a group which has only one element a such that |a| = 2. Prove that xa = ax, for all x ∈ G.

Published on 25 Sep 2014 - 5:10

score 232 · Answer 13 · 2014-09-25 16:20:25

232

Views

Can $\mathbb{Z}/n\mathbb{Z}$ (not $(\mathbb{Z}/n\mathbb{Z})^{\times}$) be a group under multiplication?

Published on 25 Sep 2014 - 16:20

score 230 · Answer 14 · 2014-09-26 06:22:07

230

Views

abelian groups?

Published on 26 Sep 2014 - 6:22

score 3.8k · Answer 15 · 2014-09-26 20:30:23

3.8k

Views

Show that an abelian group $G$ of order 55 must be cyclic.

Published on 26 Sep 2014 - 20:30

List Question

If$(ab)^n=a^nb^n$ & $(|G|, n(n-1))=1$ then $G$ is abelian

$\ker (T \otimes id_{Z})=\ker(T)\otimes Z$

Complement but not direct summand; Lam, Lectures on Modules and Rings, Example 6.17(5)

$G$ a finite group $n$-abelian goup and g.c.d.$\big(|G|,n(n-1)\big)=1$ , then to show $G$ is abelian

Direct limits of free abelian groups and diagonalization

$(\mathbb C^{\times}, \cdot)$ is a subgroup of $(GL(n,\mathbb C), \cdot)$

Groups of Order 2 with subgroups

Property of abelian groups without using Lagrange's theorem

The number, up to isomorphism, of abelian groups of order 40 is

Order of Group with Elements of Order 2

Let G be a group and a; b ∈ G. Suppose |a| = |b| = |ab| = 2. Then show that ab = ba.

Suppose G is a group which has only one element a such that |a| = 2. Prove that xa = ax, for all x ∈ G.

Can $\mathbb{Z}/n\mathbb{Z}$ (not $(\mathbb{Z}/n\mathbb{Z})^{\times}$) be a group under multiplication?

abelian groups?

Show that an abelian group $G$ of order 55 must be cyclic.

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