Share
Explore BrainMass

Subgroup proofsno

I would love to have a usual word format answer with detailed proofs.
The problems are 6 in total, the numbers are: from attached page #1: #2.82; 2.86; 2.87, from attached page #2: # 2.97; 2.98; 2.99. Thanks in advance.
The student.

Attachments

Solution Preview

Please see the attachment.

Page #1
Problem 2.82
Proof:
First, I show that is a subgroup. For any , we have and for any , then we have

Thus . So is a subgroup.
(a) I show that .
For any , we have , then for any , we can find some , such that . Now for any , we have

Therefore, . Thus .
(b) We consider the map defined as , where for any . Then we have

Thus . So is a group homomorphism.
If , then .
If , then .
Therefore, .
By the group homomorphism theorem, we have . Since is a subgroup of and , then is isomorphic to a subgroup of .

Problem 2.86
(a) " ": If is a maximal normal subgroup of , I claim that is simple. If not, we can find some proper normal subgroup in , then we have . This means that is not maximal. We get a contradiction. Therefore, is simple.
" ": If is simple, I claim that is a maximal normal subgroup of . If not, we can find some proper normal subgroup of , such that , then is a proper subgroup of . ...

Solution Summary

This is a series of proofs regarding groups and subgroups.

$2.19