1
vote
1answer
23 views

Rational group algebras and maximal orders

Let $G$ be a finite group, and $\mathbb{Q}[G]$ be the rational group algebra. Then the group ring $\mathbb{Z}[G]$ is an order in $\mathbb{Q}[G]$, but is not in general a maximal order. What can we ...
0
votes
1answer
35 views

Are the generators of the subgroup defining tensor products linearly independent over $\mathbb Z$?

Let $S$ be a (commutative) ring with identity, and let $M$, $N$ be $S$-modules. (I guess if $S$ isn't commutative, I want $M$ to be a right $S$-module an $N$ a left $S$-module.) In the definition of ...
7
votes
2answers
117 views

Is $\Bbb Q/\Bbb Z$ artinian as a $\Bbb Z$-module?

I'm confused. Is $\Bbb Q/\Bbb Z$ artinian as a $\Bbb Z$-module? We know that $\Bbb Z_{p^{\infty}} \subset \Bbb Q/\Bbb Z$ is artinian. The following argument is true or not ? $\mathbb Q / ...
5
votes
0answers
101 views

When is $K_0(i)$ an injection?

Suppose that $\mathcal A$ and $\mathcal B$ are two abelian categories such that $\mathcal A$ is a full subcategory of $\mathcal B$. If $i: \mathcal A\rightarrow\mathcal B$ is the inclusion functor, ...
4
votes
1answer
127 views

Commutative ring and its group-algebra, and abelian-group-algebra as a commutative ring.

In course of discussing the algebraic structures, one of my seniors is led quite naturally to considering the $\color{red} {geometric}$ version of following: Question: Since we could consider the ...
0
votes
2answers
69 views

Elementary Question about Torsion Subgroups

Let $G$ be an abelian group which is killed by multiplication with the integer $n\geq 1$. Let $n=a\cdot b$ with $a,b \geq 1$ and relatively prime. Denote by $G[a]$ resp. $G[b]$ the $a$-resp. ...
1
vote
1answer
312 views

Torsion subgroup

Prove that in a finitely generated abelian group $G$ the torsion subgroup is a direct summand (from Scott, Group Theory). Clearly, the torsion subgroup is normal because $G$ is abelian, so we have to ...
5
votes
2answers
519 views

A condition for a subgroup of a finitely generated free abelian group to have finite index

Let A be a free Abelian group of finite rank and B be a subgroup of A such that $A=B+pA$ for some prime number p, then how to prove $B$ is a subgroup of finite index in A? And if $A=B+pA$ holds for ...