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Showing that two groups are not isomorphic

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data
    Show that ##\langle \mathbb{R}_{2 \pi}, +_{2 \pi} \rangle## is not isomorphic to ##\langle \mathbb{R}, +\rangle##

    2. Relevant equations


    3. The attempt at a solution
    I know how to show that two groups are isomorphic: by finding an isomorphism between them. However, I am not sure how to show that there does not exist an isomorphism at all.
     
  2. jcsd
  3. Feb 20, 2017 #2

    fresh_42

    Staff: Mentor

    In general is non-existence always the hardest of all proofs. Sometimes it can be done by contradiction: assume something exists and derive a contradiction. But far more often this isn't as easy as it sounds, because the contradiction might not be as obvious or we cannot decide, whether something is really impossible or whether we're simply too stupid. A famous example is Fermat's Last Theorem, which has (rather quickly) been solved for small values of ##n## but turned out to be extremely difficult in general. Another famous example of our days is the question whether ##P \neq NP## really holds. This comes down to the question: Are there problems which are of intrinsic difficulty, or are we simply not smart enough?

    Fortunately it's easier in case of groups and isomorphisms. Here we only have to look at invariants, i.e. properties which are preserved under isomorphisms. If they are different in one group from the other, then the groups cannot be isomorphic. Now a list of possible invariants:

    1. The number of elements if one group is finite or both.
    2. You can look for elements of an order, that occurs in one but not in the other group.
    3. Commutativity.
    4. The center of a group.
    5. The commutator subgroup.
    6. The set of all normal subgroups.
    7. Solvability.
    8. Nilpotency.
    9. Centralizers and normalizers.
    10. In more advanced cases: representations.
     
  4. Feb 20, 2017 #3
    How are you defining ##\mathbb{R}_{2 \pi}## and ##+_{2 \pi}##?
     
  5. Feb 20, 2017 #4
    So, for example, if I wanted to show that the latter is not isomorphic to the former, could I just note that the equation ##z +_{2 \pi} z +_{2 \pi} z +_{2 \pi} z =0## has four solutions ##\{0,\frac{\pi}{2}, \pi, \frac{3 \pi}{2} \}##, while the equation ##x+x+x+x=0## has only one solution, ##\{ 0 \}##? And argue that if they were isomorphic then these equations should have the same number of solutions? Is this an example of an invariant you were describing?
     
  6. Feb 20, 2017 #5

    fresh_42

    Staff: Mentor

    Yes, #2, although I would formulate it in terms of group theory: ##\frac{\pi}{2} \in \langle \mathbb{R}_{2\pi}, +_{2\pi}\rangle ## is of order ##4## and ##\langle \mathbb{R},+\rangle ## has no elements of finite order. Can you show, that this implies ##\langle \mathbb{R}_{2\pi}, +_{2\pi}\rangle \ncong \langle \mathbb{R},+\rangle\; ##?
     
  7. Feb 20, 2017 #6
    Well, isomorphisms preserve order, right? So if there were an isomorphism, that would imply that ##\langle \mathbb{R},+\rangle## has at least one element of order 4. However, it has no elements of finite order, which means that there cannot exist an isomorphism.
     
  8. Feb 20, 2017 #7

    fresh_42

    Staff: Mentor

    Yes, but the "work" to be done is in the first sentence. Why do they preserve the order?

    Only as an addition to my first post (you have been too fast to edit it in time):
    Your solution basically says: I have three elements (##\frac{\pi}{2}\, , \,\pi\, , \,\frac{2\pi}{3}##) of an order, that divides ##4##, but none of such an order in ##\langle \mathbb{R},+ \rangle\, .## (The neutral ##0## aside which is automatically of order ##1\,.##)
     
  9. Feb 20, 2017 #8
    I hope this isn't too much of an imposition, but I am still interested in an answer to this question.
     
  10. Feb 20, 2017 #9

    fresh_42

    Staff: Mentor

    Addition of angles modulo ##2\pi##, I think, ##U(1)\;.##
     
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