# Show that C^x is isomorphic to GL2(R)

• November 1st 2010, 06:09 PM
kathrynmath
Show that C^x is isomorphic to GL2(R)
Show that C^x is isomorphic to the subgroup of GL2(R) consisting of all matrices of the form:
a b
-b a such that a^2+b^2 is not 0.

I'm guessing I need to show C^x--->G given by c(a+bi)=
a b
-b a is an isomorphism.

I need to show 1-1, onto, and homomorphism.
1-1:
Assume c(a+bi)=c(c+di). I need to show this implies a+bi=c+di
a b c d
-b a = -d a
I'm not sure how this implies 1-1

onto:Not sure how to show this

homomorphism:
c(ab)=c(a)c(b)
Let a=a+bi, b=c+di
a b c d
-b a * -d c

• November 1st 2010, 06:44 PM
kathrynmath
i wasn't really sure how to write the matrices so I was looking for in the first
first row: a b
second row : -b a
• November 1st 2010, 06:48 PM
Drexel28
Quote:

Originally Posted by kathrynmath
Show that C^x is isomorphic to the subgroup of GL2(R) consisting of all matrices of the form:
a b
-b a such that a^2+b^2 is not 0.

I'm guessing I need to show C^x--->G given by c(a+bi)=
a b
-b a is an isomorphism.

I need to show 1-1, onto, and homomorphism.
1-1:
Assume c(a+bi)=c(c+di). I need to show this implies a+bi=c+di
a b c d
-b a = -d a
I'm not sure how this implies 1-1

onto:Not sure how to show this

homomorphism:
c(ab)=c(a)c(b)
Let a=a+bi, b=c+di
a b c d
-b a * -d c

For injectivity remember that two matrices are equal if and only if their entries are equal.

For surjectivity it's much, much easier than you may think. Try it one more time.
• November 1st 2010, 06:56 PM
kathrynmath
so the matrices are equal if a=c, b=d

surjective we want y=c(a+bi)=first row a b
second row -b a
want to solve for a+bi
Do I let y= a b
c d
and solve that way
So c=-b, d=a
a+bi=d-ci
I feel like that was completely wrong
• November 1st 2010, 07:00 PM
Drexel28
Quote:

Originally Posted by kathrynmath
so the matrices are equal if a=c, b=d

surjective we want y=c(a+bi)=first row a b
second row -b a
want to solve for a+bi
Do I let y= a b
c d
and solve that way
So c=-b, d=a
a+bi=d-ci
I feel like that was completely wrong

What about the fact that $c\left(a+bi\right)=\begin{bmatirx}a & b\\\ -b & a\end{bmatrix}$
• November 1st 2010, 07:06 PM
kathrynmath
I don't know where that comes from I have c(a+bi)=
first row : a b
second row -b a