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anilorap

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.

  • 2 years ago
  • 2 years ago

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

    • 2 years ago
  2. anilorap
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    yea,,, I have hourssss trying to :(

    • 2 years ago
  3. badreferences
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    That's true as well for non-Abelian groups, right?

    • 2 years ago
  4. badreferences
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    Wait, yes it is. I was derping; I forgot the exact definition of Abelian for a second.

    • 2 years ago
  5. KingGeorge
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    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.

    • 2 years ago
  6. anilorap
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    I know that in order to prove the right identity i use some how my left inverse... but i keep getting stuck

    • 2 years ago
  7. badreferences
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    Man, I'm too sleepy for this. :(

    • 2 years ago
  8. anilorap
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    bad reference... please... i have 3 nights with no sleep :(

    • 2 years ago
  9. badreferences
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    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.

    • 2 years ago
  10. anilorap
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    :(

    • 2 years ago
  11. KingGeorge
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    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.

    • 2 years ago
  12. KingGeorge
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    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.

    • 2 years ago
  13. KingGeorge
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    That took some time to take apart and figure out exactly what was going on.

    • 2 years ago
  14. anilorap
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    im still trying to understand the inverses

    • 2 years ago
  15. KingGeorge
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    Any particular step in that proof?

    • 2 years ago
  16. anilorap
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    well,, im trying to understand step by step... let me see if i get stuck

    • 2 years ago
  17. badreferences
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    Woops, I tried proving it backwards using the right inverse. I feel dumb.

    • 2 years ago
  18. anilorap
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    waittt

    • 2 years ago
  19. anilorap
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    if x is the inverse of a and if we have left inverse the will be (x*a) nooo (a*x)

    • 2 years ago
  20. KingGeorge
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    If that was unclear, it means that \(x*a=e\).

    • 2 years ago
  21. KingGeorge
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    And we want to show that \(a*x=e\) as well.

    • 2 years ago
  22. anilorap
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    yea... but u are using the right one

    • 2 years ago
  23. anilorap
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    yea but you can apply that statement like its true.. u want to get that statement

    • 2 years ago
  24. KingGeorge
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    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.

    • 2 years ago
  25. anilorap
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    but my understanding is that u cant just say x*a=e=a*x.. u have to get the answer

    • 2 years ago
  26. KingGeorge
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    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 \(x*a=a*x=e\)

    • 2 years ago
  27. anilorap
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    ok let em check again

    • 2 years ago
  28. KingGeorge
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    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.

    • 2 years ago
  29. anilorap
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    ok thanks you very much I kind of getting it..

    • 2 years ago
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