Determine Whether Two Matrix Groups Are Isomorphic Or Not. 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?
 A: Assuming that $G$, $H$, and $K$ are the respective subgroups of $GL(2, \mathbb{R})$, then you can show that


*

*$H$ and $K$ are abelian, so $H \times K$ is abelian, but

*$G$ is not abelian.


The property of being abelian is preserved by any isomorphism, so the groups cannot be isomorphic.
A: (Migrated from a comment that was too long to remain a comment):
Please note, whenever you are spending A LOT OF time trying to "prove" that two groups ARE isomorphic, and find yourself not making ANY headway, then it's time to consider that perhaps they ARE NOT isomorphic, in which case no isomorphism exists...and in which case, you simply need to prove WHY NOT. If no isomorphism exist, there's no point in looking for one, once you've tried that with no success.
What structural characteristics, which any two isomorphic groups must share, exist for one group that do/does not exist for the other group? As Sammy notes in his answer, the property of being an abelian group is a structural property preserved by any isomorphism, if one exists. If one group is abelian and the other is not abelian...then...the two groups CANNOT be isomorphic: that is to say, there cannot exist any isomorphism between an abelian group and a non-abelian group.

It will be a good undertaking for you to collect notes about such structural properties. Another example, besides being commutative: if one group is cyclic, and the other is not, they cannot be isomorphic. There are many other such properties, not necessarily applicable here, but worth knowing when you suspect that two groups are not isomorphic.
