# Showing Two Groups are Isomorphic

1. Aug 3, 2004

### Oxymoron

How would I show that two groups are isomorphic?

FOR EXAMPLE:

Take the group homomorphism &phi; : ((0, oo), x) &rarr; ((0, oo), x) defined by &phi; (x) = x&sup2;

Since &phi; is taking any element in (0, oo) and operating on it by x, does it map one-to-one and onto to (0, oo)?

I assume by showing that two groups are isomorphic you have to show that there is a one-to-one correspondence and that they are onto (ie. the two groups are a bijection).

Would I start by taking some element a of ((0, oo), x) and then say that under x, a is mapped to a&sup2;. Then for all a, a&sup2; is in (0, oo) hence it is one-to-one. Then show that there is only one a that maps to a&sup2; in (0, oo) hence it is onto. Since it is both then it is isomorphic.

Any help would be appreciated.

2. Aug 3, 2004

### gravenewworld

To show a mapping is isomorphic you must show 2 things (ok really 3 things), that the mapping is is a bijection and that is preserves operation.

To show that it is 1-1 assume that P(a)=P(b) (P is phi). Prove that a=b.

To show that it is onto you must show that for any element g* in G* (G* is the group that G maps to) there exists a g in G that gets mapped to it, i.e. P(g)=g* .

Finally show that the mapping preserves operation, that is P(ab)=P(a)P(b) for all a,b in G.

3. Aug 3, 2004

### matt grime

well, you've two xs in that post that mean different things.

firstly, the map must be a homomorphism of groups, you've not shown that squaring is a homomorphism from the positive numbers under * to itself. then it suffices to show this map is a bijection which is trivial (since it has an inverse).

4. Aug 3, 2004

### Oxymoron

Matt, are you saying that if the map has an inverse then it is automatically a bijection. In other words, you can either show 1-1 and onto OR show that it has an inverse?

Here is my proof that the two groups are isomorphic...

To show that &phi; is 1-1 assume &phi;(a) = &phi;(b) for any a,b in (0,oo). Prove a = b.

Take a,b in (0, oo) then &phi;(a) = a&sup2;, &phi;(b) = b&sup2;

If &phi;(a) = &phi;(b) then a&sup2; = b&sup2; which implies that a = b.
Hence &phi; is 1-1

To show that &phi; is onto show that for any g* in (0, oo) there exists a g in (0, oo) such that &phi;(g) = g*

Take g in (0,oo) then &phi;(g) = g&sup2; = g* since then g* is going to be in (0, oo) because it is a homomorphism under multiplication, then &phi; is onto.

Finally, it preserves operation since taking any a in (0, oo)
&phi; (ab) = (ab)&sup2; = a&sup2;b&sup2; = &phi;(a)&phi;(b)

Since &phi; is 1-1 and onto then the two groups (0, oo) and (0, oo) are isomorphic.

5. Aug 3, 2004

### matt grime

a map is a bijection iff it has an inverse, that is a trivial fact, one i'm surprised you've not seen if you're doing group theory.

6. Aug 3, 2004

### Oxymoron

However, if I had a function

&phi; : (Z, +) &rarr; (Z, +) defined by &phi; (n) = n&sup2;

This is NOT homomorphic is it? Since for any a,b in Z

&phi;(a+b) = (a+b)&sup2; = a&sup2; + b&sup2; + 2ab

which does not equal &phi;(a)+&phi;(b)

Does this mean automatically that the two groups are not isomorphic?

7. Aug 3, 2004

### Muzza

Surely you mean "Take g in (0, oo) such that g = sqrt(g*)".

No, it just means that the function named phi you described is not an isomorphism. There are more functions from (Z, +) to (Z, +) than phi(n) = n^2, the proof that phi(n) = n^2 is not an isomorphism obviously says nothing about the (possible) non-existance of other functions which /are/ bijective homomorphisms (they exist though, f(n) = n will work).

Last edited: Aug 3, 2004
8. Aug 3, 2004

### Oxymoron

Matt, I am not doing group theory. I am doing linear algebra but for some reason the faculty decided to throw in a two-week crash course in elementary abstract algebra.

9. Aug 3, 2004

### Oxymoron

Muzza, correct. I omitted the next step.

10. Aug 3, 2004

### Muzza

Just FYI, the omitted step is crucial and your argument is not valid without it ;)

11. Aug 3, 2004

### Oxymoron

Muzza, if g = √(g*) then g is not in Z since Z is the set of integers. √(g*) is not always an integer. Could you explain? I thought it was isomorphic on the condition that g be back in Z?

Last edited: Aug 3, 2004
12. Aug 3, 2004

### Muzza

But we aren't in Z anyway, we're in (0, oo) which I figured was the positive reals (and all positive reals have a real positive square root). If (0, oo) denotes the positive integers (which would be very strange), then ((0, oo), *) isn't a group anyway (what would the inverse of 2 be?)...

*edit* Replaced "rationals" with "integers".

Last edited: Aug 3, 2004
13. Aug 3, 2004

### Oxymoron

R OMG! I am so stupid. I got mixed up between my two questions. Sorry about that.

Well, then that makes perfect sense then!

If we changed the set we where working on to the reals...

&phi; : (R, +) &rarr; (R, +) defined by &phi;(x) = 2x

This would be a group homomorphism because it preserves addition
For any a,b in R we have

&phi; (a+b) = 2(a+b) = 2a + 2b = &phi; (a) + &phi; (b)

And the identity is mapped back into R...

&phi; (0) = 2(0) = 0

This is also an isomorphism by a similar argument.

I think I am getting the hang of this...

14. Aug 3, 2004

### gravenewworld

Be extremely careful when working with groups. One of the first pitfalls my professor taught me is how to algebraically work with groups. The most common mistake students make when working with a group is that to do (ab)^2=a^2b^2 for a,b elements of some group G. This does not always work, although it might in this case. A better way to show that it preserves operation is by P(ab)=(ab)^2=(ab)(ab)=a(ba)b (by associativity of a group)=a(ab)b (you can permute since the group is abelian)=(aa)bb=a^2b^2 (by associative property again). I know it seems stupid to do it like this, but you have to be picky when working with groups.

15. Aug 3, 2004

### matt grime

Just because some map is not an isomorphism does not imply that the two groups are not isomorphic. The identity map between them is obviously an isomorphism in this case.

16. Aug 3, 2004

### Oxymoron

Thanks for the tips Matt and gravenewworld.

det : (GL_2(R), *) &rarr; (R \ {0}, X), where * is matrix multiplication. (X is multiplication)

Should I begin by taking two matrices A and B in GL_2(R) and show that det(A*B) = det(A)Xdet(B)?

17. Aug 3, 2004

### Muzza

I suppose that could be a good start, but you didn't say what you were asked to prove so it's kind of hard to say for sure. ;)

18. Aug 3, 2004

### matt grime

it's certainly a homomorphism presumably you've got to show that the hard way.

clearly it isn't injective, so not an isomorphism, though it is obviously surjective as well (one need only think of diagonal matrices to see why).

19. Aug 3, 2004

### Oxymoron

Sorry about that. You have to show whether or not it is a group homomorphism and decide if it is an isomorphism.

20. Aug 3, 2004

### matt grime

diag{1,x} and diag{x, 1/x}