Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anilorap

  • 2 years ago

Let (G,∗) be a binary structure that has the following properties: 1) The binary operation ∗ is associative. 2) There exists an element e∈G such that for all a∈G, e∗a=a (Existence of left Identity). 3) For all a∈G, there exists b∈G such that b∗a=e (existence of left inverses) Prove that (G,∗) is a group.

  • This Question is Closed
  1. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    All you need to show, is that it also has a right identity, and a right inverse.

  2. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea,,, I have hourssss trying to :(

  3. badreferences
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That's true as well for non-Abelian groups, right?

  4. badreferences
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wait, yes it is. I was derping; I forgot the exact definition of Abelian for a second.

  5. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    It is true for non-abelian groups too, but I'm starting to see why he was having problems. I keep getting that statements that use circular logic.

  6. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I know that in order to prove the right identity i use some how my left inverse... but i keep getting stuck

  7. badreferences
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Man, I'm too sleepy for this. :(

  8. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    bad reference... please... i have 3 nights with no sleep :(

  9. badreferences
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm running in circles, and I can't tell if I'm just falling for a trap, or if I'm just derping because I'm sleepy.

  10. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :(

  11. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    According to the always helpful wikipedia page http://en.wikipedia.org/wiki/Elementary_group_theory, Let \(y\) be the inverse of \(a*x\), and \(x\) be the inverse of \(a\). Then \[e=y*(a*x)\]\[e=y*(a*(e*x))\]\[e=y*(a*((x*a)*x))\]\[e=y*(a*(x*(a*x)))\]\[e=y*((a*x)*(a*x))\]\[e=(y*(a*x)*a)*x\]\[e=e*(a*x)\]\[e=a*x\]Thus we have a right inverse also.

  12. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    As for right identity, let \(x\) be the inverse of \(a\). Then\[a*e=a*(x*a)\]\[a*e=(a*x)*a\]\[a*e=e*a\]\[a*e=a\]So we also have right identities.

  13. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    That took some time to take apart and figure out exactly what was going on.

  14. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im still trying to understand the inverses

  15. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Any particular step in that proof?

  16. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well,, im trying to understand step by step... let me see if i get stuck

  17. badreferences
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Woops, I tried proving it backwards using the right inverse. I feel dumb.

  18. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    waittt

  19. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if x is the inverse of a and if we have left inverse the will be (x*a) nooo (a*x)

  20. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    If that was unclear, it means that \(x*a=e\).

  21. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    And we want to show that \(a*x=e\) as well.

  22. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea... but u are using the right one

  23. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea but you can apply that statement like its true.. u want to get that statement

  24. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    In the proof, using meticulous use of associativity, it shows that if \(y*(a*x)=e\), and \(x*a=e\), then \(a*x=e\). Both \(y\) and \(x\) are left inverses of \(a*x\) and \(a\) respectively.

  25. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but my understanding is that u cant just say x*a=e=a*x.. u have to get the answer

  26. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    We didn't just say that. By converting \(x\) into \(e*x\) and \(e\) into \(x*a\) (true by hypothesis), we can reorganize our terms to prove that \(x*a=a*x=e\)

  27. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok let em check again

  28. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I need to go to bed now, but if you continue to have questions, feel free to ask and I'll get back to you tomorrow.

  29. anilorap
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok thanks you very much I kind of getting it..

  30. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.