Solving Abstract Algebra Problem: Proving Isomorphism & Listing Generators

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SUMMARY

The discussion centers on proving that the function phi: Z → H, defined by phi(n) = [1 n; 0 1], is an isomorphism, where H is the subgroup of GL(2, R) defined by H = {[1 n; 0 1] | n ∈ Z}. The user struggles with the concept that [1 n; 0 1] is not a valid matrix in GL(2, R), leading to confusion about H being a subgroup. The conclusion is that phi is indeed an isomorphism, and the generators of H can be listed as the matrices corresponding to each integer n.

PREREQUISITES
  • Understanding of group theory and isomorphisms
  • Familiarity with the general linear group GL(2, R)
  • Basic knowledge of matrix operations and properties
  • Concept of subgroups in abstract algebra
NEXT STEPS
  • Study the properties of isomorphisms in group theory
  • Learn about the structure and properties of GL(2, R)
  • Explore examples of subgroups and their generators in abstract algebra
  • Review matrix multiplication and its implications in group theory
USEFUL FOR

Students and educators in abstract algebra, mathematicians interested in group theory, and anyone looking to deepen their understanding of isomorphisms and matrix groups.

WM07
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Can some one help me, how to solve this problem?. Please explain me how is done, been having problem with the subject


Let H be the subgroup of GL(2, R) under Matrix multiplication defined by
H = {[ 1 n ]}| n E Z }
0 1


Let 0: Z à H be the function defined by

phi(n) = [ 1 n ]
0 1

How do I prove phi is an isomorphism and how I list the generators

I tried to add the two matrix, but I am getting 0's, I just need explanation on the problem
 
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[1 n] is not a matrix in GL(2,R) so H clearly can not be a subgroup of GL(2,R).

Either you're missing something or I'm missing something completely.
 

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