1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Groups, monoids and nonempty subsets

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A) If M is any monois, let M' denote the set of all nonempty subsets of M and define an operation on M' by XY = {xy | x in X, y in Y}. show that M' is a monoid, commutative if M is, and find the units.

    B) If ab=ba in a monoid M, prove that (ab)^n = a^nb^n for all n >= 0.

    3. The attempt at a solution
    A) we need to show that M' is associative and has inverse.
    E = {e} in M' -- identity in M'
    X in M' XE = {xe | x in X} = {x | x inX} = X so that EX = X
    It seems right, but I have no idea how to show associative..
    Z in M' X(YZ) = {x(yz) | x in X, y in Y, and z in Z} = (XY)Z = { (xy)z |x in X, y in Y, and z in Z} is it right for associative??
    find all units in M'
    If a in M -- a unit in M
    {a} in M' -- a unit in M
    {a}{a^-1} = {aa^-1} = {e} = E
    {a}{a^-1} = {e} = E ----------I think it's correct..

    B)By induction. If n = 1, (ab)^1 = a^1b^1 which is true. If k >= 1, we assume that 1,2,...,k are all true, (ab)^k = a^kb^k. we must show that k+1 is also true.
    (ab)^k+1 = (ab)^k * ab
    = a^kb^k * ab
    = a^k (b^k * a)b
    = a^k(a*b^k)b = (a^k*a)(b^k*b) = a^k+1 b^k+1 complete induction.
    I think it's correct, but it was wrong somewhere..
    I don't know where..

    Thank you.
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted

Similar Discussions: Groups, monoids and nonempty subsets
  1. Subsets and Groups (Replies: 1)

  2. Nonempty subset homework (Replies: 18)