Here is the problem I am having trouble with:
Let:
$$G = \left\{\begin{bmatrix} a & b\\ 0 & c \end{bmatrix} \in GL(2,R)\right\}$$
$$H = \left\{\begin{bmatrix} a & 0\\ 0 & b \end{bmatrix} \in GL(2,R)\right\}$$
$$K = \left\{\begin{bmatrix} 1 & a\\ 0 & 1 \end{bmatrix} \in GL(2,R)\right\}$$
Is $G$ isomorphic to $H \times K$?
Let's say that:
$\phi : G \rightarrow H \times K$ where:
$\begin{bmatrix} a & b\\ 0 & c \end{bmatrix} \rightarrow (\begin{bmatrix} a & 0\\ 0 & b \end{bmatrix}, \begin{bmatrix} 1 & a\\ 0 & 1 \end{bmatrix})$
Then, this weird mapping clearly isn't the bijection. However, I've only shown that one function isn't an isomorphism. The thing is: I need to find the bijective homomorphism, which are injective and surjective.
Some notes I made are:
- G consists of any upper triangular matrices, including the identity matrix.
- H x K is the Cartesian product of two matrices. That is:
$(\begin{bmatrix} a & 0\\ 0 & b \end{bmatrix}, \begin{bmatrix} 1 & a\\ 0 & 1 \end{bmatrix})$. But this is a set of H x K. If I multiply the same types of matrices with any arbitrary constants coordinate-wise, then I obtain the group: $(\begin{bmatrix} aa' & 0\\ 0 & bb' \end{bmatrix}, \begin{bmatrix} 1 & a + a'\\ 0 & 1 \end{bmatrix})$ - A function is homomorphism if $\phi (xy) = \phi (x) \phi (y)$
- A function is injective if its kernel consists of only an identity element.
- The image of the inverse of the element is congruent to the preimage of the same element. That is: $\phi (x^{-1}) = \phi^{-1}(x)$
The problem is that I don't know where to start off for this problem. I can't seem to find the bijection.
Any advices or comments?
Best Answer
Assuming that $G$, $H$, and $K$ are the respective subgroups of $GL(2, \mathbb{R})$, then you can show that
The property of being abelian is preserved by any isomorphism, so the groups cannot be isomorphic.