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Algebra Theorem

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Let H be a subset of a group G. Prove: H does not equal the empty set and a,b are contained in H, which implies ab^(-1) is contained in H, which implies H is a subgroup.

    Let G be a group. Let Ha = {x is contained in G | ax = xa }. Prove Ha is a subgroup of G
    .
    2. Relevant equations

    none

    3. The attempt at a solution

    I looked at these and had issues figuring out where to start. If someone could give me somewhat of an idea of a starting point I could probably work it all out. I generally find this to be the biggest predicament for me: finding out where to start.
     
  2. jcsd
  3. Feb 12, 2007 #2

    matt grime

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    What are the definitions of subgroup? Show you can deduce these from the rule:

    for all a,b in H, then ab^-1 is in H.

    Let's get you started: you need to show that for all b in H, that b^-1 is in H for H to be a subgroup. Now, if e (the identity) were in H, and b were in H how would you show that b^-1 is in H from the rule above?

    Can you show e is in H?
     
  4. Feb 12, 2007 #3

    Dick

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    Ok, where to start? The assertion is that these two sets are groups. To be a group you have to satisfy certain properties (identity, closure etc). So list these properties and then start proving them for these sets. There. You are started.
     
  5. Feb 12, 2007 #4
    I couldn't type it above, but for the first one, I also have to prove: If H is a subgroup, then ab^(-1) is contained in H. The final arrow should actually be pointing in both directions, if that makes sense.

    Isn't it just assumed that e is always contained in H? I see exactly where you are going with it to a degree. The definition of group is that it's closed, it has an identity, and it has an inverse. Therefore, it's just assumed from the fact that its a group that e exists.
     
  6. Feb 12, 2007 #5

    Dick

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    Do you really think you can 'assume it to be a group' if you are trying to 'prove that it is a group'?
     
  7. Feb 12, 2007 #6
    It was actually a given above if you read where it says "Let H be a subset of group G." After all of that, I have to prove what follows.
     
  8. Feb 12, 2007 #7

    Dick

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    Let's do one thing at a time, ok? You seem to be trying to prove things with 'the final arrow' running both ways at once. Start with H is a SUBSET such that a*b^(-1) is in H for any a,b in H. Tell me how to show e is in H.
     
  9. Feb 12, 2007 #8

    HallsofIvy

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    This isn't at all clear. Do you mean IF H does not equal the empty set and WHENEVER a,b are contained in H THEN ab-1 is contained in H, implies that H is a subgroup?

    What was given by that? The empty set is subset of G. A set containing a single member of G is a subset of G. That alone implies nothing.

    Certainly just knowing that "H is a subset of
     
  10. Feb 12, 2007 #9

    matt grime

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    <breathe, take a step back>

    1. H is a subgroup of G if H, g in H implies g^-1 in H, and g,h in H implies gh in H (closed under inverse and composition).

    2. H is a subgroup of G if for all g,h in H gh^-1 is in H.


    Exercise: show these two definitions are equivalent.

    This is what we're trying to show. 1=>2 trivially, and 2=>1 equally trivially with a couple of insights about how to choose g and h.
     
  11. Feb 13, 2007 #10

    HallsofIvy

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    Notice that in "if a, b, are in H then ab-1 is in H", it is NOT required that a and b be different. What happens if, for a in H, you look at a, a instead of a, b?
     
  12. Feb 13, 2007 #11
    Ok, I have all of #1 proved from varying ideas that were scrambling around in my mind overnight. I'm having some problems finding closure in the last problem, however. If someone could assist me in that, it would be fantastic!!

    #1 - Someone correct me if I am wrong!!
    To prove identity, I simply said a = x and b = x, therefore, ab^(-1) = xx^(-1) = e. Therefore, we have the identity.

    Inverse would simply be setting a = e, b^-1 = x^-1, then ab^(-1) = ex^(-1), which of course would simply be x^(-1). Therefore, inverse exists.

    Closure would be if we set a = x, b = x, then ab^(-1) = e. Since e is contained in H, H is closed.

    Going backwards from the given that H is a subgroup and trying to prove that H is not empty, all we need to do is show that it has only a single property of a subgroup, in which case the empty set wouldn't exist. I hope my logic is correct. Anyway, I stated that H has an identity, therefore, for any a,b contained in H, there exists an e contained in H such that aa^(-1) = e. Therefore, e is contained in H, therefore H does not equal the empty set.


    #2 - Corrections would be nice!!
    Identity:
    Since x is contained in G
    There exists an x^(-1) contained in G, and there exists and e contained in G, since this is part of the definition of a group.

    Suppose x = e contained in G,
    Since ax = xa, we know ae = ea, which implies a = a, therefore e is contained in H.

    Inverse:
    Suppose x = a^(-1)
    Since ax = xa, that implies that aa^(-1) = a^(-1)a which implies that e = e. Therefore, an inverse exists contained in H.

    Closure is where I'm stuck, so if anyone can give me a hand with that, then it would be greatly appreciated!!
     
  13. Feb 13, 2007 #12

    matt grime

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    #1, what has that got to do with closure? Closure: x in H, u in H imples xy is in H.

    I have no idea what you're doing in #2.

    If H is a subgroup, with the normal definition, it is trivial that x,y in H implies xy^-1 is in H. There is bugger all to prove in that direction.
     
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