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Couple questions involving monoids/isomorphisms

  1. Sep 20, 2006 #1
    I'm having trouble with these two questions.

    Problem 1: Prove that the multiplicative monoid N of natural numbers 1, 2,... is a free commutative monoid.

    Problem 2: Is the multiplicative monoid N isomorphic to the additive monoid N_0 x N_0 x ...x N_0 (n times), for any n = 1, 2,...? Prove your claim.

    If i can get any tips on where to begin with them, that would be great. from there i will work on an show where i have progressed. right now, all i can think of is using induction in 2 to prove that there is an isomorphism for all n. i do this, by looking at neutral elements in each monoid, and constructing the isomorphism for n=1. that's all i can think up.
    Last edited: Sep 20, 2006
  2. jcsd
  3. Sep 20, 2006 #2
    i actually figured out 1. it was fairly simple. 2 is a little trickier though. haven't made any progress on it, im afraid.
  4. Sep 20, 2006 #3
    heres what i've been thinking:
    every value in N can be written as a product of primes, which can be directly related to N_0 x N_0 x ...x N_0. for instance 45= 9 * 5= (3^2)*5. this can be written as (2,1,0,0,...). where the values in that "vector" are the powers of increasing powers of primes. my problem comes when i start thinking of when n=1. what occurs then? i know the concept is there, and that i might end up using properties of commutative free monoids, but what?
  5. Sep 21, 2006 #4

    matt grime

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    Can something that is not finitely generated be isomorphic to something that is?
  6. Sep 21, 2006 #5
    no, which is exactly the case with n an integer. so is that already my solution? that i should prove there is no isomorphism? if so, then does it suffice to use that as a proof? that (N_0,+) which itself has rank 1 (and further more any monoid with rank n) cannot be isomorphic to a monoid with countably infinte generators? i believe that's a theorem in itself - two monoids are isomorphic if their ranks are equal
    Last edited: Sep 21, 2006
  7. Sep 21, 2006 #6
    one thing that troubles me, however, is that i have the following equivalent statements in my notes. perhaps i haven't full understood its meaning?

    "the following are equivalent for a monoid S:

    i) S =~ F_Abmon (X) for some X

    ii) there exists an X, submonoid of S such that for all a in S, a not equal to e_S (netural in S), there exist a unique representation a= * x_i ^k_i for i=1 to r (r,k_i are greater or equal to 1, and no x_i are similar."

    here, * is the concatination of elements. ii) resembles problem 2 in the sense that every element in N can be written as a product of primes, the generators in N. this would mean that the isomorphism would exist between (N_O,+) and (N,x)....according to ii), anyway. this is exactly the opposite of what you just said, i believe, since their ranks are totally different.

    edit: i should note, just in case, that F-Abmon (X) is just the free commutative monoid of X, and =~ means isomorphic.
    Last edited: Sep 21, 2006
  8. Sep 21, 2006 #7

    matt grime

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    Why would you think that ii) implies an isomorphism of monoids between N_0 underr addition and N under multiplication? Since * in the former is addition and * in the latter is multiplication there is no isomorphism: there is no uniqueness about writing an integer as the *sum* of primes (5=5=2+3).
  9. Sep 21, 2006 #8
    that cleared some things up. thanks.

    that being said, does it suffice to say that since N_0 X N_0 X ... X N_0 (n times) equipped with addition , for any n=1,2,3,... is generated by n values, and (N,*) has countably infinite generators that they cannot be isomorphic? or is there a more extensive proof of this?
  10. Sep 21, 2006 #9

    matt grime

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    Of course there is a more extensive proof - you could write out all the details. What level of detail you need to furnish depends on for whom you are writing.
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