Prove GL(R) is not isomorphic to GL(C)

  • Thread starter ArcanaNoir
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In summary: Since \phi(-\text{I})=\pm \text{I}, and \phi(i\text{I})^2=\phi(-\text{I}), and \phi(i\text{I}) is a matrix of real numbers, it cannot be that \phi(-\text{I})=-\text{I}.This implies \phi(-\text{I})=\pm \text{I}, which shows that \\phi is not injective.You seem to be saying that if we can find a matrix A such that ##A^2=I##, then the equation ##phi(-\text{I})=-\text{I}##
  • #1
ArcanaNoir
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Homework Statement



Prove GL(R) is not isomorphic to GL(C)

Homework Equations





The Attempt at a Solution



Well I don't really have a good grasp of the issue at hand as this is another appendix problem, but my thoughts are I think GL(R) is a subgroup of GL(C) so maybe I could show they don't have the same cardinality? Would that be enough? Been a while since I did algebra... Not sure how I would would show that either.
 
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  • #2
Do you know why ##\mathbb{R}\setminus \{0\}## is not isomorphic to ##\mathbb{C}\setminus \{0\}##? Try to generalize.
 
  • #3
micromass said:
Do you know why ##\mathbb{R}\setminus \{0\}## is not isomorphic to ##\mathbb{C}\setminus \{0\}##? Try to generalize.

Umm [tex] f(I)=f(I^2)=f(I)^2[/tex] implies f(I) is the identity in GL(C). [STRIKE]But [tex]f(-I)^2=f(I)[/tex] so f(-I) is also the identity in GL(C). [/STRIKE]

Is that how to do it?

Errr. Flaw in that logic. meh.
 
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  • #4
Okay here's what I got. Comments are appreciated.

Suppose there exists an isomorphism [itex] \phi : \text{GL}(\mathbb{C}) \rightarrow \text{GL}(\mathbb{R})[/itex]. Then [itex]\phi(\text{I})=\text{I}[/itex]. We also have [itex]\phi (-\text{I})^2=\phi((-\text{I})^2)=\phi(\text{I})=\text{I}[/itex]. This implies [itex]\phi(-\text{I})=\pm \text{I}[/itex]. Since [itex]\phi(i\text{I})^2=\phi(-\text{I})[/itex], and [itex]\phi(i\text{I})[/itex] is a matrix of real numbers, it cannot be that [itex]\phi(-\text{I})=-\text{I}[/itex]. Hence, [itex]\phi(-\text{I})=\text{I}[/itex], which shows that \[itex]\phi [/itex] is not injective.
 
  • #5
Why does ##A^2 = I## imply that ##A=\pm I##? And why can ##A^2 = -I## not occur?
 
  • #6
micromass said:
Why does ##A^2 = I## imply that ##A=\pm I##?


Because the square root of I is ±I right?



And why can ##A^2 = -I## not occur?

Because real numbers squared cannot be negative 1.

Right?
 
  • #7
ArcanaNoir said:
Because the square root of I is ±I right?





Because real numbers squared cannot be negative 1.

Right?

You seem to assume that these things hold for matrices because they hold for real numbers. You need to prove whether they also hold for arbitrary matrices.
 
  • #9
Office_Shredder said:

Alas!

micromass said:
You seem to assume that these things hold for matrices because they hold for real numbers. You need to prove whether they also hold for arbitrary matrices.

Well... Do they hold? Seems like they don't based on the link Office_Shredder provided.

Your original direction was to show it holds for numbers, so if you pretend I wrote 1's instead of I's, that's my logic. Now I just need to get this stuff to generalize. Did you have any hint on how to do that?
 
  • #10
Maybe you can find the centers of both groups?
 
  • #11
micromass said:
Maybe you can find the centers of both groups?

That would be any scalar (except for 0) times the identity matrix?
 
  • #12
ArcanaNoir said:
That would be any scalar (except for 0) times the identity matrix?

Yes. Are the centers isomorphic?
 
  • #13
micromass said:
Yes. Are the centers isomorphic?

I would guess not. I would like to say because of my proof, but I know I'm missing something.
 
  • #14
ArcanaNoir said:
I would guess not. I would like to say because of my proof, but I know I'm missing something.

Following up on mm's suggestion, can you think of a group that is isomorphic with a real non-zero scalar times the identity matrix?
How about a group that is isomorphic with a complex non-zero scalar times the identity matrix?I also have an alternative solution method: what you can say about the eigenvalues of the matrix A if ##A^2=I##?Btw, I'm somewhat confused about the meaning of ##GL(\mathbb R)##.
What does it mean?
I suspect you do not mean what wiki says about the General linear group.
 
  • #15
I like Serena said:
Following up on mm's suggestion, can you think of a group that is isomorphic with a real non-zero scalar times the identity matrix?
How about a group that is isomorphic with a complex non-zero scalar times the identity matrix?

Reals minus 0 with multiplication?

Btw, I'm somewhat confused about the meaning of ##GL(\mathbb R)##.
What does it mean?
I suspect you do not mean what wiki says about the General linear group.

Either nxn invertible matrices or infinite ones, not sure what my professor intended. If I can get it figured out for nxn I'm confident it will generalize.
 
  • #16
ArcanaNoir said:
Reals minus 0 with multiplication?

Yep.

micromass said:
Do you know why ##\mathbb{R}\setminus \{0\}## is not isomorphic to ##\mathbb{C}\setminus \{0\}##? Try to generalize.

Now consider this previous comment of mm...
 

FAQ: Prove GL(R) is not isomorphic to GL(C)

What is GL(R) and GL(C)?

GL(R) refers to the general linear group of real numbers, which is the set of all invertible square matrices with real number entries. GL(C) refers to the general linear group of complex numbers, which is the set of all invertible square matrices with complex number entries.

What does it mean for two groups to be isomorphic?

Two groups are isomorphic if there exists a bijective function between them that preserves the group operation. In simpler terms, this means that the two groups have the same structure and behave in the same way.

How can we prove that GL(R) is not isomorphic to GL(C)?

We can prove this by showing that there does not exist a bijective function between GL(R) and GL(C) that preserves the group operation. This can be done by examining the properties of these two groups and showing that they are not equivalent.

What are the key differences between GL(R) and GL(C)?

The key difference between GL(R) and GL(C) lies in the properties of their elements. In GL(R), the elements are real numbers and the group operation is multiplication. In GL(C), the elements are complex numbers and the group operation is also multiplication. However, complex numbers have additional properties such as having an imaginary component and obeying different rules of arithmetic compared to real numbers.

Are there any other ways to prove that GL(R) is not isomorphic to GL(C)?

Yes, there are other ways to prove this such as using group theory concepts like order and subgroups, or by examining the cardinality (size) of the two groups. Additionally, we can also prove this by showing that certain transformations or operations on one group do not have corresponding counterparts in the other group.

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