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Abstract algebra

  1. Sep 6, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    I am asked to show that if E is a semi-group and if

    (i) there is a left identity in E
    (ii) there is a left inverse to every element of E

    then, E is a group.


    3. The attempt at a solution
    Well I can't seem to find the solution, but it's very easy if one of the two "left" above is replaced by a "right". For instance, if we replace the existence of a left inverse condition by the existence of a right inverse, then we find that the left identity is also a right identity like so:

    Let a,b be in E. Then ab=a(eb)=(ae)b ==> a=ae (by multiplying by bˉ¹ from the right). So e is a right identity also. Then it follows that every right inverse is also a left inverse:

    aaˉ¹=e ==>(aaˉ¹)a=ea ==>a(aˉ¹a)=a ==> (aˉ¹a)=e.

    So, does anyone know for a fact that this question contains or does not contain a typo?
     
  2. jcsd
  3. Sep 6, 2007 #2

    quasar987

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    No, this only means that ae=ae. :frown:
     
  4. Sep 6, 2007 #3

    Hurkyl

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    Well, I can prove that all of the left inverses of e are, in fact, equal to e. So I'm making progress. :smile:
     
  5. Sep 6, 2007 #4

    Hurkyl

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    Also,

    x * (left inverse of x) = (left identity)

    I think that the group structure follows from these facts. So I have at least as much confidence in the original problem as I do that I didn't make a mistake. (I'm not saying how much confidence that is. :wink:)
     
  6. Sep 6, 2007 #5

    Dick

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    I'm having problems with this as well, but then I'm tired. I would suggest though, that if you really think it's wrong that you start trying to construct a counterexample. If you can't construct a counterexample then the effort may teach you what you need to do.
     
    Last edited: Sep 6, 2007
  7. Sep 6, 2007 #6
    I know that I've done this proof before, and as I recall you need to somehow use the two facts in conjuction, and remember that if y is the left inverse of x, then y also has a left inverse say z... I remember having to use this someow, but my efforts on this tonight are not going well.

    Edit: With a fair amount of work I managed to prove that every left inverse is also a right inverse, and from that I think it follows a little more easily that the left identity is also a right identity.

    Edit 2: It actually follows almost trivially from the fact that every left inverse is also a right inverse that the left identit is also the right identity.
     
    Last edited: Sep 6, 2007
  8. Sep 6, 2007 #7

    Hurkyl

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    Basically, I tried writing lots of expressions that could be simplified in multiple ways to derive new properties. For example, I first wondered how left inverses of the indentity (and their left inverses, etc) behaved, then I started worrying about how inverses of general elements behaved.

    Incidentially, I did get started by searching for a counterexample; I decided to let 0 be the identity, 1 a left inverse of 0, 2 a left inverse of 1, and so forth, then I tried to compute how multiplication had to behave.
     
  9. Sep 7, 2007 #8
    I think I got it. According to my calculation, the statement is true as stated. But since I'm not a mathematician and did in fact not even know the terms left-inverse and left-identity before tackling the problem (I found left-inverse in an algebra book, didn't find left-identity) I need some sanity-check:
    I assumed the two conditions mean that:

    [tex] \exists \, 1_L \in E : \forall a \in E : 1_L \, a = a[/tex] (i)
    and
    [tex] \forall \, a \in E: \, \exists \, a_L \in E: a_L \, a = 1_L [/tex] (ii).
    Is this translation of the two conditions correct?
     
    Last edited: Sep 7, 2007
  10. Sep 7, 2007 #9

    matt grime

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    It is, though it is always preferable not to write things in logical forms like that since they are unnecessarily opaque.
     
  11. Sep 7, 2007 #10
    i guess this is pretty easy:
    i would go like this:

    From knowing E is a semi group you have associativity!
    And i know:

    e*a=a , for e beeing a left identity!

    and

    a^-1 * a = e , a^-1 is left inverse

    now i multiply a from the left and get:

    a*a^(-1)*a = a

    from using associativity i get:

    (a*a^(-1))*a = a*(a^(-1)*a) = a (1)

    hence by using both assumptions the first part of (1) implies that a*a^(-1) = e -> every left inverse is a right inverse if associativity holds!

    the second part implies that while we assumes a^(-1)*a= e -> every left identity is a right identity.

    QED
     
  12. Sep 7, 2007 #11
    I don't completely understand what you said Mr. Brown. Most notably, I don't get the step
    which seems to imply that a*e=a.
     
  13. Sep 7, 2007 #12

    quasar987

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    You went a little too fast. Multiplying from the left by a gives

    a*a^(-1)*a = ae

    but ae is not known to be a, for e is only a left identity.
     
  14. Sep 7, 2007 #13

    quasar987

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    Let me try something here. (Assuming the problem is stated correctly)

    If I show that (aˉ¹)ˉ¹=a, then this will mean that e=(aˉ¹)ˉ¹(aˉ¹)=aaˉ¹, meaning left inverses are also right inverses.

    Lets begin the random manipulations :)

    (aˉ¹)ˉ¹(aˉ¹)=e ==>(aˉ¹)ˉ¹(aˉ¹)a=ea ==> (aˉ¹)ˉ¹e=a ==> aˉ¹(aˉ¹)ˉ¹e=aˉ¹a ==>aˉ¹(aˉ¹)ˉ¹e=e ==> aˉ¹(aˉ¹)ˉ¹=e ==> aˉ¹=((aˉ¹)ˉ¹)ˉ¹.

    If every element can be seen as the left inverse of another, then I have succeeded. But is this implied? Gotta go.
     
  15. Sep 7, 2007 #14
    ae=a strikes back.
    I don't think it's a good idea to label the left-inverse and left-identity [tex]a^{-1}[/tex] and e/1. That nomenclature imho cries for stupid mistakes caused by that you usually label real inverses and identities with these symbols. Might differ from person to person, but I did chose different names for exactly the reason that I screwed up too many steps otherwise.
     
    Last edited: Sep 7, 2007
  16. Sep 7, 2007 #15

    Hurkyl

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    How'd you manage that? You have neither proven that e is a right identity, that e has a unique left inverse, or anything else I've noticed that would allow you to conclude that.
     
  17. Sep 7, 2007 #16

    quasar987

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    I accepted w/o proof that

    :smile:
     
  18. Sep 7, 2007 #17

    learningphysics

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    I was stuck trying to figure out this same problem recently. I looked it up online... I can't find the link right now.

    But the trick is to first prove:

    If x*x = x, then x = e, for any element x.

    This is simple

    x*x = x
    x^-1 * x * x = x^-1 * x (left multiply both sides by x^-1)
    e * x = e
    x = e

    Then you can show that every left inverse is a right inverse

    x*x^-1
    = x * (e * x^-1)
    = x * (x^-1 * x) * x^-1 (write out e as x^-1*x)
    = (x * x^-1) * (x * x^-1)

    So using the previous solution we know that x * x^-1 = e, so the right inverse part is proven

    So to prove e is a right identity:
    x * e
    = x * (x^-1 * x)
    = (x * x^-1) * x
    = e * x
    = x

    Then you can also prove that e is the unique left identity and unique right identity.
     
  19. Sep 7, 2007 #18

    quasar987

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    Cheers!
     
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