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Order of Join and Intersection of Subgroups

  1. Jun 27, 2012 #1
    1. The problem statement, all variables and given/known data
    If S and T are subgroups of a finite group G, prove that
    [S:1][T:1] ≤ [S[itex]\cap[/itex]T:1][S[itex]\vee[/itex]T:1]


    2. Relevant equations
    Notation: [A:B] is the number of cosets of B in A for some subgroup B of A.
    note that [A:1] is the order of A.

    Lagrange's Theorem:
    For some subgroup S of G.
    [G:1]=[G:S][S:1]



    3. The attempt at a solution
    [S:1] ≤[S[itex]\vee[/itex]T:1]
    and many other statements like this... none of them seem to help much.
     
  2. jcsd
  3. Jun 27, 2012 #2

    jbunniii

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    What can you say about the order of the set

    [tex]ST = \{st | s \in S, t \in T\}[/tex]?

    Can you relate this to the orders that appear in the desired inequality?
     
  4. Jun 28, 2012 #3
    |ST| = Sum of orders of distinct cosets of S of the form St for some element t.

    For two cosets St and St', we have that St=St' iff St*t'[itex]^{-1}[/itex]=S, which means [itex]t*t'^{-1}\in S[/itex]. But [itex]t*t'^{-1}\in T[/itex] also, so we must have [itex]t*t'^{-1}\in (S\cap T)[/itex]. Then [itex](S\cap T)t=(S\cap T)t'[/itex] so that there are the same number of cosets St in ST ([ST:S]) as there are cosets [itex](S\cap T)t\ of\ (S\cap T)\ in\ T \ \ [T:(S\cap T)][/itex]. Applying Lagrange's theorem twice:

    [itex][ST:S][/itex]=[itex][T:(S\cap T)][/itex] implies:

    [itex]\frac{|ST|}{|S|}=\frac{|T|}{|S\cap T|}[/itex], so that

    [itex]|ST|=\frac{|S||T|}{|S\cap T|}[/itex] and since clearly [itex]ST\subset(S\vee T)[/itex]

    [itex]\frac{|S||T|}{|S\cap T|}=|ST|≤|S\vee T|[/itex], so

    [itex]|S||T|≤|S\vee T||S\cap T|[/itex]

    I got the proof that [itex][ST:S][/itex]=[itex][T:(S\cap T)][/itex] from Dummit and Foote - this is not in the book I'm working through (MacLane and Birkhoff), which is where this problem is from. Is this supposed to be obvious? I don't know how I would have ever done this problem without just happening upon this result. Thanks!
     
  5. Jun 28, 2012 #4

    jbunniii

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    Your solution looks good. The key result you used, [itex][ST : S] = [T : (S \cap T)][/itex] is a standard one which most books either prove or assign as an exercise (so at least the statement appears in the book). I guess having that result makes your problem quite easy, so perhaps MacLane and Birkhoff expected you to discover it for yourself, or perhaps to solve the problem in some other way. Are you using their elementary text (A Survey of Modern Algebra) or the more advanced one (Algebra)?

    Do they prove the 2nd ismorphism theorem? Namely, if N is a normal subgroup and H is any subgroup, then [itex]NH/N[/itex] is isomorphic to [itex]H/(N \cap N)[/itex]. This implies the order result you used in the case where one subgroup is normal. From this, it's not a major stretch to conjecture that the order result might be true even if neither subgroup is normal.
     
  6. Jun 28, 2012 #5
    Hi Jbunniii,

    Thanks for all your help! I'm using Algebra (the advanced one) - as my primary text as I work through abstract algebra for the first time, but I've been reading parts of a few other algebra textbooks (Dummit+Foote, Fraleigh, Artin, and some course notes from a prof at uc berkeley). I'm studying the subject on my own, which is proving to be a bit insane, though this forum helps.

    They don't prove any isomorphism theorems until 300 pages after this problem (in a chapter called "The Structure of Groups"). Due to your hint about ST, I was reminded that I had seen reference to such a set in D+F, which turned out to be right next to the section on the isomorphism theorems. In MacLane and Birkhoff, the definitions of kernel, image, and normal subgroup don't even appear until the next section!

    The only thing I can think of is that they expect you to contemplate the nature of |S∨T| until you get the idea to consider |ST:S|?

    Thanks again for your help, I feel I'd have been stuck on this forever.
     
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