# Showing Two Groups are Isomorphic

by Oxymoron
Tags: groups, isomorphic, showing
 P: 867 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.
 P: 1,408 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.
 Sci Advisor HW Helper P: 9,396 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).
 P: 867 Showing Two Groups are Isomorphic 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.
 Sci Advisor HW Helper P: 9,396 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.
 P: 867 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?
P: 695
 Quote by Oxymoron 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).
 P: 867 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.
 P: 867 Muzza, correct. I omitted the next step.
P: 695
 Muzza, correct. I omitted the next step.
Just FYI, the omitted step is crucial and your argument is not valid without it ;)
 P: 867 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?
 P: 695 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".
 P: 867 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...
 P: 1,408 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.