Here is the problem I am having trouble with:


$$G = \left\{\begin{bmatrix} a & b\\ 0 & c \end{bmatrix} \in GL(2,R)\right\}$$

$$H = \left\{\begin{bmatrix} a & 0\\ 0 & b \end{bmatrix} \in GL(2,R)\right\}$$

$$K = \left\{\begin{bmatrix} 1 & a\\ 0 & 1 \end{bmatrix} \in GL(2,R)\right\}$$

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?

  • 2
    $\begingroup$ There are some basic terminology and notation issues with your question. What you have written, e.g. $$G=\begin{bmatrix} a & b\\ 0 & c \end{bmatrix} \in GL(2,R)$$ says that $G$ is a matrix, not that $G$ is the group of such matrices. Additionally, two groups can be isomorphic to each other, and a function can be an isomorphism, but it does not make sense to ask whether a function is isomorphic. $\endgroup$ – Zev Chonoles Apr 26 '13 at 23:14
  • $\begingroup$ Thanks for letting me know that. I was not careful about the notation I use. I though that I need to type with {}, but they don't show up. Instead, I include these \left\{ \right\} $\endgroup$ – NasuSama Apr 26 '13 at 23:18
  • 1
    $\begingroup$ Braces are used for grouping in $\LaTeX$-something between them is treated as one object. For example, to get multidigit powers, you use x^{12} to get $x^{12}$. To get braces to display, you need to escape them with a backslash, so \{ gives $\{$, but don't need the \left and \right. Those will make the braces big enough to enclose whatever they are around. $\endgroup$ – Ross Millikan Apr 26 '13 at 23:22
  • 1
    $\begingroup$ Just a quick note on your last bullet point. The notation $\phi(x^{-1}) = \phi^{-1}(x)$ is probably better written as $\phi(x^{-1}) = \phi(x)^{-1}$. The point is that the RHS is not a preimage, but another inverse. You can summarize it as "The image of the inverse of [an] element is [equal to] to inverse of the image of the same element." $\endgroup$ – Jason DeVito Apr 26 '13 at 23:30

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.

  • $\begingroup$ Can you fill in the proofs of the bulleted claims? $\endgroup$ – Sammy Black Apr 26 '13 at 23:19
  • $\begingroup$ I bet he can. The question is why won't you prove them...or prove them false, of course? $\endgroup$ – DonAntonio Apr 26 '13 at 23:30
  • $\begingroup$ I have proven that if H and K are abelian, then H x K is abelian. Now, I need to show that G is not abelian. In order to show this, do I need to consider the inverse of any homomorphism, or is this not considered? $\endgroup$ – NasuSama Apr 26 '13 at 23:45
  • 1
    $\begingroup$ Never mind. I've prove that G is not abelian while H x K is abelian. Sammy's response really makes sense to me. $\endgroup$ – NasuSama Apr 26 '13 at 23:55
  • $\begingroup$ Good, Michael! It is indeed a great answer. $\endgroup$ – Namaste Apr 26 '13 at 23:58

(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.

  • $\begingroup$ I haven't learned much about group properties in my class until I joined this site and asked the question. Very informative site! $\endgroup$ – NasuSama Apr 26 '13 at 23:58
  • $\begingroup$ Lot's to learn...we're all learning! Stick around! And good job with showing some of your thoughts and knowledge about the problem, as well as your efforts to format! $\endgroup$ – Namaste Apr 26 '13 at 23:58
  • $\begingroup$ @amWhy: Deserves an uptick from me! + 1 $\endgroup$ – Amzoti Apr 27 '13 at 0:24
  • $\begingroup$ And this is the final opening door to get that $15$. I assume this is an small gift from me to you these days. And I am so happy that I could be the first one who did it for making you a bit enlightened Amy. :-) $\endgroup$ – mrs Dec 31 '13 at 7:55
  • $\begingroup$ @amWhy: Nothing happens. :-( $\endgroup$ – mrs Dec 31 '13 at 8:03

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