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Is 2Z isomorphic to 4Z? (Abstract algebra)

  1. Apr 3, 2012 #1
    Actually I'm stupid today, it happens once in a while that I get extremely lazy and stupid in mathematics, but today I came up with a bizarre thing in abstract algebra that I couldn't find my mistake on my own and I'm not sure whether what I've concluded is true or wrong, I was proving another theorem that one of my assumptions implied: [itex](\Bbb{2Z},+)\cong (\Bbb{4Z},+)[/itex]. This is the story:
    Consider the map [itex] f: 2\Bbb{Z} \to 4\Bbb{Z}[/itex] defined by f(n)=2n. It's obvious that f is surjective and it's easy to verify that f is injective because if f(m)=f(n) then 2m=2n and we obtain: m=n. This map is a homomorphism of groups because:
    f(m+n)=2(m+n)=2m+2n=f(m)+f(n)
    so it implies that [itex](\Bbb{2Z},+)\cong (\Bbb{4Z},+)[/itex]. Is that really true? I guess it must be but it sounds a bit weird to me. This result implies that every infinite cyclic group has a subgroup of it which is isomorphic to itself!!! sounds weird and counter-intuitive to me :( Although I must admit that it's just as weird as N and Z having the same cardinal numbers so maybe it's OK? Anyway, is it true that every infinite cyclic group has a subgroup of it which is isomorphic to itself?
     
  2. jcsd
  3. Apr 3, 2012 #2
    Consider that every infinite cyclic group is isomorphic to the integers, and that isomorphism is an equivalence relation.
    If an infinite cyclic group has an infinite cyclic subgroup, then they must be isomorphic (by transitivity); now just prove that every infinite cyclic group has such a subgroup. You might try using the fact (as you're doing here) than the set of all multiples of an integer n is an infinite subgroup, and since isomorphism maps subgroups to subgroups...
     
    Last edited: Apr 3, 2012
  4. Apr 3, 2012 #3
    Actually proving that every infinite cyclic group has such a group is a bit hard, I could prove it in an easier way.

    If G is cyclic, then all elements of G can be generated by a specific element like a. so we have G=<a> iff g = an for any g in G. Now, every subgroup of G is cyclic too, this can be proved easily, therefore there must be a power of a that generates S (S being a subgroup of G). so the only thing I need to do is to map a to am where m is the power of a that generates the subgroup and everything is just like before in integers.
    It's very easy to verify that f: G → S defined by f(a)=am is an isomorphism.

    So the statement is true?
     
  5. Apr 3, 2012 #4

    I like Serena

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    Yep.

    Any infinite cyclic group G=<a> is isomorphic to <a2>, which is a subgroup.
    Same proof as the one you already gave.
     
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