Is there an isomorphism from G to (G/M)x(G/N) with the kernel M\bigcapN?

  • Context: Graduate 
  • Thread starter Thread starter symbol0
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on proving the isomorphism G/(M∩N) ≅ (G/M) × (G/N) where M and N are normal subgroups of G and G = MN. The user attempts to establish a homomorphism f(a) = (aM, aN) with the kernel M∩N, leveraging the fundamental homomorphism theorem. The challenge lies in demonstrating that this homomorphism is onto, which requires showing that for any elements g and g' in G, there exists an element a in G such that (gM, g'N) = (aM, aN). The discussion highlights the necessity of M and N being normal subgroups for the proof to hold.

PREREQUISITES
  • Understanding of group theory and isomorphisms
  • Familiarity with normal subgroups and their properties
  • Knowledge of the fundamental homomorphism theorem
  • Experience with constructing homomorphisms in algebraic structures
NEXT STEPS
  • Study the properties of normal subgroups in group theory
  • Learn about the fundamental homomorphism theorem in detail
  • Explore examples of isomorphisms in finite groups
  • Investigate the construction of homomorphisms and their kernels
USEFUL FOR

This discussion is beneficial for mathematicians, particularly those specializing in abstract algebra, group theorists, and students seeking to deepen their understanding of group isomorphisms and normal subgroups.

symbol0
Messages
77
Reaction score
0
Let M and N be normal subgroups of G such that G=MN.
Prove that G/(M\bigcapN)\cong(G/M)x(G/N).

I tried coming up with an isomorphism from G to (G/M)x(G/N) such that the kernel is M\bigcapN, so that I can use the fundamental homomorphism theorem.
I tried f(a) = (aM, aN). It is an homomorphism and M\bigcapN is the kernel but I'm having a hard time showing it is onto.

I would appreciate any help.
Thank you
 
Physics news on Phys.org
So you need to show that for all g, g' in G, there is an a in G such that: (g M, g' N) = (a M, a N).

I haven't worked this out in detail, but: since G = MN, you can write g = m n, g' = m' n'. I suspect that a = m' n might do the trick.
You will need that M and N are normal, so in particular h M = M h, h N = N h for all h in G.
 
Thank you Compuchip
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
683
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 26 ·
Replies
26
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
4
Views
2K