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  1. M

    Generating Functions!

    Thanks,Sir...that was a big help :-)
  2. M

    Generating Functions!

    That's precisely what my professor said in class!Can you give me examples where they're used?
  3. M

    Generating Functions!

    I meant f(x) is that expression and not (f(x))^-1. But well...I just got to know that the problem can be solved not using ordinary generating functions but will require "something" called exponential generating functions. Can Someone here please give me a motivation for the concept of...
  4. M

    Generating Functions!

    We are about to begin this topic soon in class... I'd like to know all about generating functions and their application. I tried reading it up...One thing I'd like to know is, once you have a generating function...then what? You get some information...but what is it? I tried to find no. of...
  5. M

    Subgroup math help

    have been thinking about prob 3....i guess you take A6 (the set of even permutations of 6 elements), possibly find a subgroup H of order 4 and then look at the 3 distinct cosets of H.
  6. M

    Groups again

    suppose G is abelian and has 2 elements of order 2, say a and b...then {e,a,b,ab} becomes a subgroup of G. but 4 does not divide 2p unless p=2. if p=2 then anyway {e,a} is a subgroup of G of order 2( p in this case).
  7. M

    More group theory!

    i guess i must( or rather i am supposed to) add my thought process... since 3 does not divide n (n,3)=1 so we have un for any x in G, x = x^(un+3v)= x^3v. so G ={y^3 / y is in G}...using this i think we're supposed to show that ab=ba for all a in G. here's where i'm stuck.
  8. M

    More group theory!

    seeing lots of group theory here after a really long time... let G be a finite group of order n, where n is not divisible by 3. suppose (ab)^3 = a^3 b^3 ,for a, b in G . prove that G is abelian.
  9. M

    Irreducible elements

    Here’s an interesting question… Let R be a commutative ring and ‘a’ an element in R. If the principal ideal Ra is a maximal ideal of R then show that ‘a’ is an irreducible element. If a is prime, this is pretty obvious…if a is not prime, then we say a= bc for some b,c in R. Now we need to show...
  10. M

    Euclidean domains

    this seems to be a very fundamental problem...but i need help... prove or disprove : let D be a euclidean domain with size function d, then for a,b in D, b != 0, there exist unique q,r in D such that a= qb+r where r=0 or d(r) < d(b). first of all, what is size function? we only...
  11. M

    Idempotent elements am + bn= 1 implies......1-bn = am that implies... bn(1-bn) = abmn = 0... so bn is an idempotent element .... i looked at " am +bn =1" a hundred times before posting this question....but it flashed just now! thanks a ton!
  12. M

    Idempotent elements

    here's a real tough one ( at least for me) that the ring Z/mnZ where m ,n are relatively prime has an idempotent element other than 0 and 1. i looked at examples and it works.... do we look for solutions of the equation a^2 -a = kmn , for some k in Z( that is, other than 0 and 1)? help!
  13. M

    Finite fields

    thanks...that makes it clear... but here's another question :redface: does this mean i could have done this problem without considering separate cases???
  14. M

    Finite fields

    let F= { 0=a1, a2,a3,} be a finite field. show that (1+a2)(1+a3).........(1+an) = 0. when n is odd, it's simple since 1 belongs to F. then odd number of elements are left( they're distinct from 0 and 1). at least one of them, say x,must have itself as its inverse. x^2 = 1...
  15. M

    What a homeomorphism means

    hi! would like to know what a homeomorphism means ( how do you geometrically visualize it?) AND is the symbol 8 homeomorphic to the symbol X? Why or why not? ( from whatever little i know intuitively about homeomorphisms, i think it is not....)
  16. M

    Permutation groups

    i think the solution 504 is correct.... i don't see any fallacy in it.
  17. M

    Eigenvalues problem

    Yeah, you use cayley-hamilton theorem so, you have p(A)=0... That implies a*A^n + b*A^(n-1) + c*A^(n-2)........+ I= 0 (i've used a,b,c as coefficients)...then take the identity matrix to the other side. Multiply both sides by inverse of A. Then RHS becomes -A^(-1) the LHS shows that the...
  18. M

    Permutation groups

    hey galileo please read stuff carefully... we left this a long while ago, right astronut?! :smile: