Showing Two Groups are Isomorphic

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In summary, to show that two groups are isomorphic, one must demonstrate that the mapping is a bijection and preserves operations. In the conversation, the example of a group homomorphism φ : ((0, oo), x) → ((0, oo), x) defined by φ (x) = x² is used. The conversation also discusses the steps necessary to prove that this mapping is both one-to-one and onto, and therefore isomorphic. It is important to be careful when working with groups, as common algebraic operations may not always work in this context.
  • #1
Oxymoron
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How would I show that two groups are isomorphic?

FOR EXAMPLE:

Take the group homomorphism φ : ((0, oo), x) → ((0, oo), x) defined by φ (x) = x²

Since φ 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 a². Then for all a, a² is in (0, oo) hence it is one-to-one. Then show that there is only one a that maps to a² in (0, oo) hence it is onto. Since it is both then it is isomorphic.

Any help would be appreciated.
 
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  • #2
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
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
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 φ is 1-1 assume φ(a) = φ(b) for any a,b in (0,oo). Prove a = b.

Take a,b in (0, oo) then φ(a) = a², φ(b) = b²

If φ(a) = φ(b) then a² = b² which implies that a = b.
Hence φ is 1-1

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

Take g in (0,oo) then φ(g) = g² = g* since then g* is going to be in (0, oo) because it is a homomorphism under multiplication, then φ is onto.

Finally, it preserves operation since taking any a in (0, oo)
φ (ab) = (ab)² = a²b² = φ(a)φ(b)

Since φ is 1-1 and onto then the two groups (0, oo) and (0, oo) are isomorphic.
 
  • #5
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
However, if I had a function

φ : (Z, +) → (Z, +) defined by φ (n) = n²

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

φ(a+b) = (a+b)² = a² + b² + 2ab

which does not equal φ(a)+φ(b)

Does this mean automatically that the two groups are not isomorphic?
 
  • #7
Oxymoron said:
To show that φ is onto show that for any g* in (0, oo) there exists a g in (0, oo) such that φ(g) = g*

Take g in (0,oo) then φ(g) = g² = g* since then g* is going to be in (0, oo) because it is a homomorphism under multiplication, then φ is onto.

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

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

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).
 
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  • #8
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
Muzza, correct. I omitted the next step.
 
  • #10
Muzza, correct. I omitted the next step.

Just FYI, the omitted step is crucial and your argument is not valid without it ;)
 
  • #11
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?
 
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  • #12
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".
 
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  • #13
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...

φ : (R, +) → (R, +) defined by φ(x) = 2x

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

φ (a+b) = 2(a+b) = 2a + 2b = φ (a) + φ (b)

And the identity is mapped back into R...

φ (0) = 2(0) = 0

This is also an isomorphism by a similar argument.

I think I am getting the hang of this...
 
  • #14
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
Oxymoron said:
However, if I had a function

φ : (Z, +) → (Z, +) defined by φ (n) = n²

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

φ(a+b) = (a+b)² = a² + b² + 2ab

which does not equal φ(a)+φ(b)

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

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
Thanks for the tips Matt and gravenewworld.

This one had me stumped...

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

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
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
Sorry about that. You have to show whether or not it is a group homomorphism and decide if it is an isomorphism.
 
  • #20
diag{1,x} and diag{x, 1/x}
 
  • #21
Matt, how did you see that it isn't injective? I am having some difficulty working this out :-C
 
  • #22
what's the determinant of a diagonal matrix with entries x and 1/x for any non-zero x?
 
  • #23
The determinant of a diagonal matrix is just the product of the entries on the main diagonal. So a 2x2 diagonal matrix with diagonal entries x is x². And the determinant of the 2x2 diagonal matrix with entries 1/x is 1/x². Is this what you mean?
 
  • #24
The determinant of a 2x2 diagonal matrix with elements x and 1/x (x non-zero) would be 1, not 1/x^2... I.e. there are many matrices which are not equal to one another, but have the same determinant.

Also, you can prove that no isomorphism can exist between those two groups ((R\{0}, x) is abelian but GL_2(R) isn't).
 
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  • #25
I see, I misunderstood. The 2x2 diagonal matrix has entry(11)= x and entry(22) = 1/x with entries(12,21) = 0. Then I agree, the determinant is 1.

But, as you said, many matrices have determinant 1 so the mapping between the two groups is not bijective. Hence it is not isomorphic? You said that because one group is abelian and the other is not indicates that the two groups are not isomorphic, is this another property that I do not know about? Could you explain.
 
  • #26
But, as you said, many matrices have determinant 1 so the mapping between the two groups is not bijective. Hence it is not isomorphic?

No, "hence det(x) is not an isomorphism between the two groups" (this might be what you meant though). There is a difference between saying that two groups are isomorphic and saying that a particular function is an isomorphism between two groups.

You said that because one group is abelian and the other is not indicates that the two groups are not isomorphic, is this another property that I do not know about?

Try constructing the proof yourself (proof by contradiction).
 
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  • #27
Why did you choose the diagonal entries to be x and 1/x?

I understand that the identity is mapped to itself
det(I) = 1
And that the group operation is preserved. I can show this by proving
det(A*B) = detA X detB not equal to 0.
Hence the det is a group homomorphism between the two groups.

Thanks for the help guys.
 
  • #28
I chose x and 1/x because that was the most obvious way of showing that the map is not injective. Always make your life easier on yourself; diagonal and upper triangular matrices have determinants and traces that are easy to read off.
 
  • #29
I have another quick question.

If I were asked to prove that a certain operation * was a binary operation on a given set S. Would I simply prove that it obeys closure? That is, take x, y in S and show that x*y is back in S. If so, then it is a binary operator on S.
 
  • #30
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  • #31
So if my set S was the set of all real numbers excluding -1. And (S, *) was a group where x*y = x + y + xy. How would I start proving that * is a binary operation?
 
  • #32
you would show it satisfied all the axioms defining a binary operation, that's all. It's a "just do it" proof. Clearly given two inputs there is a unique output, how about closure? Note, since you've called (S,*) a group, then * must be a binary operation, or it isn't a group.
 
  • #33
The hints here have been outstanding. I wanted to add something though, and I can only think of this summary of what has been said:

1. to show two groups are isomorphic, you must find an isomorphism between them,

2. to show two groups are not isomorphic, you must show there cannot be any isomorphism, not just that one particular attempt fails. this is harder, as your argument has to apply to all potential isomorphisms. hence you need to find a property of groups that would be preserved by all isomorphisms, and yet which your two groups do not share, such as being commutative (which is called "abelian" for groups, in honor of Niels Abel).

In general the search for properties that would be preserved by all isomorphisms is a deep and fundamental one in every area, sometimes called the search for "invariants".

for example in algebraic curve theory, to show the projective plane curve x^3 + y^3 = z^3, is not rationally isomorphic to the line, can be done by outright cleverness, but is most efficiently done by producing the invariant called the genus. I.e. topoloogically the cubic is a doughnut and the "line" is a sphere.

the proof of the fundamental theorem of algebra in topology, is done by finding some way of discerning the difference between the punctured plane and the plane itself, which eventually becomes the first homology group. i.e. you have to show why the unit circle cannot be pulled away from the origin without passing through the origin. this is usually done by computing the integral of dtheta, and applying greens theorem from calculus.

in number theory one uses reduction "mod n" which says that any solution of an equation in integers would also yield a solution mod every n. Hence, since after division by 4, the equation x^2 = 2 has no solution (the left side always has remainder 0 or 1 after division by 4,) hence the equation x^2 = 204,840,962 also has no solution.
 

1. How do you prove two groups are isomorphic?

The most common way to prove that two groups are isomorphic is by finding a bijective homomorphism between them. This means finding a function that preserves the group structure and is one-to-one and onto.

2. What is the significance of proving two groups are isomorphic?

Proving that two groups are isomorphic shows that they have the same underlying structure and are essentially the same group. This allows us to use properties and theorems from one group to understand the other group.

3. Can two groups have different elements and still be isomorphic?

Yes, two groups can have different elements and still be isomorphic as long as the group structure is preserved. This means that the operation and identity element are the same in both groups, and the same relationships between elements hold.

4. Is there a unique way to prove two groups are isomorphic?

No, there are multiple ways to prove two groups are isomorphic. Some common methods include finding a bijective homomorphism, showing that the groups have the same Cayley table, or using the First Isomorphism Theorem.

5. Can two non-abelian groups be isomorphic?

Yes, two non-abelian groups can be isomorphic. The abelian property is not a requirement for isomorphism, as long as the group structure is preserved.

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