MHB Finite vs Ring Groups: Examining Theorems

cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
Dear Everyone,

Does every theorem that holds for finite group holds for ring groups? Why or Why not?Thanks
Cbarker1
 
Physics news on Phys.org
What do you mean by “ring groups”?
 
A group ring defined as the following from Dummit and Foote:

Fix a commutative ring $R$ with identity $1\ne0$ and let $G=\{g_{1},g_{2},g_{3},...,g_{n}\}$ be any finite group with group operation written multiplicatively. A group ring, $RG$, of $G$ with coefficients in $R$ to be the set of all formal sum

$a_1g_1+a_2g_2+\cdots+a_ng_n$, $a_i\in R$, $1\le i\le n$
 
Well, note that $G$ is a group whereas $RG$ is a ring, so not every theorem about $G$ may be applicable to $RG$. For example, $G$ may be a cyclic group, but there is no such thing as a cyclic ring, so a theorem about cyclic groups may not make sense when applied to rings.

What you can say is that $RG$ contains a subring isomorphic to $R$, namely
$$\{a\cdot e_G:a\in R\}$$
as well as a subset which, with respect to multiplication, forms a group isomorphic to $G$, namely
$$\{1_R\cdot g:g\in G\}.$$
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
9K
Replies
1
Views
369
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
733
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K