Question

For the below pair of group show that they are not isomorphic
or construct explicitly the isomorphisms between them:
(R,+) and the set of 2x2 matrices
(1 x
0 1) where x is a real number, under matrix multiplication

Answer 1

What's your intuitive sense on this?

Answer 2

They are both infinite sets so I guess they are not isomorphic

Answer 3

Two infinite groups can be isomorphic, so that's not a good argument.

Answer 4

Let me ask you, what is the product of two arbitrary members of the matrix group?

Answer 5

I.e., what is the product of
1 x
0 1

and

1 y
0 1

Answer 6

1 x+y
0 1

Answer 7

This group theory is so confusing for me :(

Answer 8

Yes, that matrix multiplication is correct. And that suggests there might be a very natural isomorphism between (R,+) and this matrix group, call it (M, .)

Answer 9

It looks just like addition but one is just a number the other is a matrix. So I can see something common but dont really get it

Answer 10

Am I right that we are searching for a function that maps from one group to the other? But how is that possible between 2 different things?

Answer 11

Define a function
\[ f : (\mathbb{R},+) \rightarrow (M, .) \]
by \[f(x) = \left[\begin{matrix}1 & x \\ 0 & 1\end{matrix}\right]\]

Answer 12

Now you just need to prove that is an isomorphism.

Answer 13

Things like that it is closed?

Answer 14

Actually, first you need to show that it is in fact a group homomorphism:
i.e., f(x + y) = f(x).f(y), f(0) = 0

Answer 15

and then you need to show it is 1:1 and onto

Answer 16

"Things like it is closed" is proved that the group operation doesn't take two elements of a set to something outside the set, so not a property of group homomorphisms.

Answer 17

Yes, Im just reading my notes, that was something else.
I found this: have to have the same size (that is cool)
Have to have the same order. (what is the order of these?)

Answer 18

Both have infinite order, because for every non-identity member x in (R,+) and m in M, there is no integer n such that nx = 0 or nm = 0.

Answer 19

Those are necessary conditions, but not sufficient.

Answer 20

I found it!
Has to be a bijection and
alfa(x)alfa(y)=alfa(xy)

Answer 21

OK I will give it a go to prove the bijection but Im absolutely not sure if I can do it

Answer 22

right, that's the function I wrote down. note that the group operation in (R,+) is addition, +; but in (M, .), the group operation is matrix multiplication.

So you need to show that f(x+y) = f(x)f(y), where f(x)f(y) is the product of the two matrices f(x), f(y).

Answer 23

What I did:
f(x)=f(y) that is
(1 x = 1 y
0 1 0 1
Implies that x=y so its 1:1
f(x+y)=(1 x+y = (1 x ( 1 y = f(x) f(y)
0 1 ) 0 1) 0 1)

Answer 24

What do I do with the onto part?

Answer 25

You need to show that for an arbitrary member m of M, there is a g in R such that f(g) = m.

Answer 26

but that is trivial

Answer 27

Indeed. :-)

Answer 28

Thanks a lot, I have a few more of these. I will struggle with them for a while and if Im stuck I post a new question

Answer 29

ok