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I've been stuck on this proof for a while. H being normal in G

  1. Nov 21, 2014 #1
    • Member warned about posting homework questions in non-homework forum sections
    Let H subgroup in G. Prove H is normal in G iff for all g in G and for all h in H, there exists an h1 and h2 in H such that hg=gh1 and gh=h2g.


    Completely lost on this. All I know for normal is that gH=Hg. Where else do I go from here? Help a brother out! Thanks :)
     
  2. jcsd
  3. Nov 21, 2014 #2

    mathwonk

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    try writing out what gH = Hg means in terms of elements of H.
     
  4. Nov 21, 2014 #3
    So {ge, gh1, gh2,..., ghk}
    {eg, h1g, h2g,..., hkg}

    Right?

    Looking at that, I am still not making a connection. :/
     
  5. Nov 21, 2014 #4

    Stephen Tashi

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    One can try to prove things in a "slick" way or try to prove them in a knock-down-drag-out way. A slick proof usually depends on using theorems previously given in the course materials. A knock-down-drag-out-way is usually needed when the purpose of the problem is to give the student a workout in basic definitions.

    This following is about how to start a knock-down-drag-out proof:

    The statement to be proven has the form "H is normal in G if and only if ...some other stuff". So you must prove both:
    "If H is normal in G then...the other stuff" and "If the other stuff...then H is normal in G".

    A common way to prove statements of the form "If A then B" is to assume A is true and show B is true.

    To tackle "If H is normal in G then ...the other stuff", you can begin by saying "Assume H is normal in G." (You said that "H is normal in G" means gH = Hg. I think you meant to say it means "for each elment g,in G, gH = Hg". )

    The other stuff has the form "for each.... there exists....". A common way to prove a "for each...." statement is to invent a variable that represents a "specific example, but one with no special properties". (logicians calls this method "universal generalization".) So you can say "Let g be an arbitrary element of G and let h be an arbitrary element of H".

    Half of what you must prove is "there exists an element h1 of H such that hg = gh1". This is where we go from pure logic to mathematics. You need to use the fact that the definition of "H is normal in G" implies that Hg = gH for the particular g we are considering. For the particular h we are considering hg is an element of Hg (by definition of the set Hg) The equality of the sets Hg and gH implies that hg is also an element of gH. Since hg is an element of gH, it must be expressible as g(h0) for some element h0 in H. (i.e. "there exists an element h0 of H such that hg = gh0 ") So we have hg = gh0. But if you give h0 the name "h1" in your writing, you have proven hg = gh1, so do that..

    Does that get you started?

    In many math courses you aren't expected to say what principles of logic your are using, you just use them and expect the reader to accept the techniques. So your writing probably doesn't have to be a wordy as the above. It does have to explain when you use a definition or theorem.
     
  6. Nov 21, 2014 #5

    Fredrik

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    You want to prove that the following statements are equivalent:

    (a) gH=Hg.
    (b) For all ##g\in G## and all ##h\in H##, there exist ##h_1,h_2\in H## such that ##hg=gh_1## and ##gh=h_2g##.

    You can prove the equivalence ##(a)\Leftrightarrow(b)## by proving the two implications ##(a)\Rightarrow (b)## and ##(b)\Rightarrow (a)##.

    ##(a)\Rightarrow (b):## Let ##g\in G## be arbitrary. Let ##x\in gH## be arbitrary. (Now what does (a) tell you that can you say about x?)

    ##(b)\Rightarrow(a):## Let ##g\in G## be arbitrary. Let ##x\in gH## be arbitrary. (Now what does (b) tell you that you can you say about x?)
     
    Last edited: Nov 21, 2014
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