For G=GL(2,R), for f, show that f is not an inner automorphism

  • Thread starter Thread starter mathnerd1
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on proving that the function f defined by f(A)=(A^tr)^-1 for A in GL(2,R) is not an inner automorphism. The key conclusion is that there does not exist a fixed matrix B in GL(2,R) such that f(A)=BAB^-1 for all A in GL(2,R). The center of GL(2,R), denoted Z(G), consists of matrices of the form aI, where a is a scalar and I is the identity matrix. The restriction of f to Z(G) does not yield a valid inner automorphism, as demonstrated by the failure of various attempts to find suitable matrices B.

PREREQUISITES
  • Understanding of GL(2,R) and its properties
  • Knowledge of inner automorphisms and their definitions
  • Familiarity with the concept of the center of a group, Z(G)
  • Basic linear algebra, including matrix transposition and inversion
NEXT STEPS
  • Study the properties of inner automorphisms in group theory
  • Explore the structure and properties of the center of GL(2,R)
  • Investigate the implications of matrix transposition and inversion in linear transformations
  • Learn about the relationship between automorphisms and group actions
USEFUL FOR

This discussion is beneficial for students of abstract algebra, particularly those studying group theory, as well as mathematicians interested in the properties of linear groups and automorphisms.

mathnerd1
Messages
2
Reaction score
0
1. Homework Statement :

Let G=GL(2,R) and define f:G->G by f(A)=(A^tr)^-1, where A^tr represents the transpose of A for all A in GL(2,R). Prove that f is not an inner automorphism, hence prove that there does not exist a fixed matrix B in G such that f(A)=BAB^-1 for all A in G

2. Homework Equations :

Before doing this question, we were asked to find the Z(G) [centre of G=GL(2,R)].
I know that an inner automorphism on G is given by as stated in the question f(A)=BAB^-1

3. The Attempt at a Solution :

So my attempt at the solution was first to find the centre of GL(2,R)
Hence Z(G) is given by the set of matrices of the form aI, where a is a scalar and I is the 2x2 identity.
My teacher had told us a hint and said that for us to solve this question we would have to look at the restriction of f to the centre of G=GL(2,R).
After obtaining Z(G), I can't see the restriction that f places on Z(G).
I thought of the zero matrix for B, then obviously that would not be an inner automorphism because all forms of BAB^-1 for A in GL(2,R) would equal the zero matrix. But then it hit me that the zero matrix is not in GL(2,R) so that solution attempt failed.
Then I thought what about a matrix of the form where the first row was zeros and then the bottom row was scalars, but then those matrices do not belong in GL(2,R) since their determinant is 0.
I also tried multiplying arbitrary matrices, but that also got me no where.

Please help me : )
Thank you so much!
 
Physics news on Phys.org
Answer this question: what are you trying to gain by trying out several values of B? Indeed the ones you tried are not in GL(2,R), so won't work. But the point is to show that f(A) is not equal to BAB-1 for all B in GL(2,R). Proving it isn't so for some B won't work.

Your teacher's advice is good. Let's say that g(A) = BAB-1 is an inner automorphism. If M is in the center of GL(2,R), what is g(M)?
 
mmm

if g(M) for M in the centre of Z(G), that implies that M=BMB^-1 or that MB=BM?
 
Ok, so g(M)=M for M in the center where g is an inner automorphism. Does f(M)=M for all M in the center?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K