# Groups Isomorphic or Not?

1. Apr 24, 2013

### NasuSama

1. The problem statement, all variables and given/known data

Let:

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

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

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

Is $G$ isomorphic to $H x K$?

2. Relevant equations

Definition of Isomorphism

3. The attempt at a solution

In order for $G \approx H x K$, there exists an isomorphism from $G$ to $H x K$. In order for the mapping to be injective, its kernel must consist of only identity element. The inverse of the function must also exist in order for the mapping to be surjective. Oh! That function needs to be homomorphism.

I let:

$\phi : G \rightarrow H x K$
$\begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \rightarrow ( \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} , \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} )$

I would say that $G$ is not isomorphic to $H x K$ since kernel of the mapping is empty. If we take a = c = 1 and b = 0, then this gives:

$( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} , \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} )$

Taking the identity matrix for the mapping doesn't give identity matrices as shown here. No matter what values of a, b and c we take, we don't get the identity matrices.

So G is not isomorphic to H x K?

2. Apr 24, 2013

### Fredrik

Staff Emeritus
You have only shown that there's one function that isn't an isomorphism.

The $\phi$ you chose is pretty strange, because there's no c on the right-hand side of the definition. This makes it impossible for the function to be a bijection (which an isomorphism must be). If you want to prove that the groups are isomorphic (I don't know if they are), you should first find a bijection, and then check if it's an isomorphism.

You should post your definition of the group you denoted by $H\times K$. What is the multiplication operation of that group?

LaTeX tip: \times, \otimes $\times,\otimes$

3. Apr 25, 2013

### NasuSama

$H \times K$ is the Cartesian product of H and K. It's denoted by:

$(\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}, \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix})$

I am not sure how to find the bijection for this type of problem..

4. Apr 25, 2013

### Fredrik

Staff Emeritus
Yes, but what is the multiplication operation? What is the product of two arbitrary members of $H\times K$? An isomorphism is a bijective homomorphism from a group into a group, but you have defined $H\times K$ as a set, not as a group.

5. Apr 25, 2013

### NasuSama

It's something that you multiply the pairs of matrices component-wise to get the group. For instance:

$(a,a') \cdot (b,b') = (ab,a'b')$

But we don't know what is a' and b'. I'm thinking carefully about this problem when you said that I formed the set, not a group (which is indeed right!).

However, I'm thinking of the multiplication of two matrices. But we have the Cartesian product. Not sure where to start off.

6. Apr 25, 2013

### Fredrik

Staff Emeritus
This equality is the end of a statement that begins with "For all $a,b\in H$ and all $a',b'\in K$".

It's easier to keep things straight in your head if you use a more suggestive notation like (h,k)(h',k')=(hh',kk').

If
$$h=\begin{bmatrix}a & 0\\ 0 & b\end{bmatrix},\quad k=\begin{bmatrix}1 & c\\ 0 & 1\end{bmatrix},\quad h'=\begin{bmatrix}a' & 0\\ 0 & b'\end{bmatrix},\quad k'=\begin{bmatrix}1 & c'\\ 0 & 1\end{bmatrix},$$ what is (h,k)(h',k')?

7. Apr 25, 2013

### NasuSama

By multiplying the matrices coordinate-wise, I obtain:

$(\begin{bmatrix} a & ac \\ 0 & b \end{bmatrix}, \begin{bmatrix} a' & a'c' \\ 0 & b' \end{bmatrix})$

8. Apr 25, 2013

### Fredrik

Staff Emeritus

(It looks like you computed (hk,h'k') instead of (hh',kk')).

9. Apr 25, 2013

### NasuSama

Oops. Sorry about that. I rushed through the computation. I attempt to do it in my head, but I have the wrong one! :)

$(\begin{bmatrix} aa' & 0 \\ 0 & bb' \end{bmatrix}, \begin{bmatrix} 1 & c + c' \\ 0 & 1 \end{bmatrix})$

10. Apr 25, 2013

### Fredrik

Staff Emeritus
Right, so now you're looking for a bijection $\phi:G\to H\times K$ such that $\phi(xy)=\phi(x)(y)$ for all $x,y\in G$. Ignore that condition for a while. Can you think of any bijection at all?

Last edited: Apr 25, 2013
11. Apr 26, 2013

### NasuSama

I am not sure if this works. Would it be the identity homomorphism?

12. Apr 26, 2013

### NasuSama

Or probably I'm guessing for no good. In order for the homomorphism to be the bijection, it needs to be surjective and injective.

I think the identity homomorphism is a bijection (I mean any matrix mapped into the identity element regardless of the variables a, b and c.). Would there be more isomorphism besides of that?

This is getting my head spinning around since this question requires me to think. :|

13. Apr 26, 2013

### Fredrik

Staff Emeritus
It should be pretty easy to find a bijection. The first thing I wrote down that made any sense to me turned out to be a bijection. (It turned out to not be a homomorphism, but let's not worry about that just yet). Can you write down any $\phi:G\to H\times K$ at all?

It helps to consider this: What does a typical member of G look like? What does a typical member of H×K look like?

A function that takes everything to the identity is a homomorphism, but such a function is obviously not a surjection unless the codomain is a trivial group (i.e. a group with only one element).

14. Apr 26, 2013

### NasuSama

The member of G, the set, would look like the 2 by 2 upper triangular matrix.
The member of H x K would look like the "point" with two diagonal matrices.

I guess I am confused with the variables a, b and c my instructor used for the problem. :\

15. Apr 26, 2013

### NasuSama

I believe this is a bijection.

$\begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \rightarrow (\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix}, \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix})$

Last edited: Apr 26, 2013
16. Apr 26, 2013

### Fredrik

Staff Emeritus
See, that wasn't so hard was it? This is the function I had in mind when I said that the first thing I wrote down turned out to be a bijection.

I could immediately see that the function you wrote down in post #1 isn't a bijection, because it didn't mention c on the right. This made it impossible for it to be injective.

17. Apr 26, 2013

### NasuSama

But how do you know it's surjective?

18. Apr 26, 2013

### Fredrik

Staff Emeritus
It's a straightforward application of the definition of "surjective". Do you know that definition?