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P primary group and the correspondence theorem

  1. Jun 17, 2012 #1

    I have a question from "A first course in abstract algebra" by J. Rotman,

    Hi, this is a question from " A first course in abstract algebra" by J. Rotman
    define d(G) = dim(G/pG)

    chapter 5, lemma 5.8 (P392),

    Let G be a finite p primary abelian group.
    If S<=G, then d(G/S) <= d(G)

    The first line of the proof read like,

    By the correspondence theorem, p(G/S) = (pG +S)/S,

    How is this equation derived? As the correspondence theorem mainly states isomorphism, I cannot see where there is equation involved? It would be greatly appreciated if anyone could help on this. Many thanks!
  2. jcsd
  3. Jun 17, 2012 #2

    In my book (3rd edition), it is chapter 6, lemma 6.10 (i).

    Now, [itex]\,p(G/S)\,:=\{p(x+S)=px +S\;|\;x\in G\}\leq G/S[/itex] , and since this is a subgroup of the quotient [itex]\,G/S\,[/itex], the

    correspondence theorem tells us that there exists [itex]\,H\leq G\,\,s.t.\,\,p(G/S)=H/S\,[/itex] , and it's not hard to realize that in fact

    [itex]\,H=pG+S\,[/itex] , for example [itex]\,\forall x\in G\,\,,\, px + S\in p\left(G/S\right)\Longrightarrow pG+s\leq p(G/S)\,[/itex]. Now you try to prove the other way around.


    Ps The correspondence theorem is *not* about isomorphisms merely but about a 1-1 correspondence between

    subgroups of [itex]\,G/N\,[/itex] and subgroups of [itex]G\,[/itex] containing [itex]\,N\,[/itex].
  4. Jun 17, 2012 #3
    Thanks a lot DonAntonio,

    this is very helpful!

    I am just wondering
    [itex]\,p(G/S)\,:=\{p(x+S)=px +S\;|\;x\in G\}[/itex] is this a definition for p(G/S), not
    [itex]\,p(G/S)\,:=\{p(x+S)=px +pS = px+ S\;|\;x\in G\}[/itex] (i assume := means definition?)
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