Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with gr.group-theory
Search options not deleted
user 2083
Questions about the branch of algebra that deals with groups.
54
votes
Accepted
Do all exact $1 \to A \to A \times B \to B \to 1$ split for finite groups?
This is true (1). It was extended to finitely generated profinite groups here (2). Surprisingly, it is also true in the category of finitely generated modules over a Noetherian commutative ring (3).
…
- 30.6k
16
votes
The finite subgroups of SL(2,C)
Dolgachev has a note on the McKay correspondence in dimension $2$. It has a lot of cool stuff on subgroups of $SL(2,\mathbb C)$, mostly from the algebraic geometry point of view.
- 30.6k
14
votes
Zero divisor conjecture and idempotent conjecture
Clearly one implies the other as $x^2=x$ means $x(x-1)=0$.
I doubt they are known to be equivalent since the sources I found: the K-theory handbook and Alain Valette survey (see Conjecture 2) listed …
- 30.6k
1
vote
1
answer
895
views
Torsion-free and torsionless abelian groups
This question is motivated by my most spectacular answer on MO (:
Let $A$ be a module over $\mathbb Z$. $A$ is said to be torsion-free if $na=0$ implies $n=0$ or $a=0$ for any $n\in \mathbb Z, a\i …
- 30.6k