## anilorap 3 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.

1. KingGeorge

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

2. anilorap

yea,,, I have hourssss trying to :(

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

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

5. KingGeorge

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

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

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

8. anilorap

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

:(

11. KingGeorge

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

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

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

14. anilorap

im still trying to understand the inverses

15. KingGeorge

Any particular step in that proof?

16. anilorap

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

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

18. anilorap

waittt

19. anilorap

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

20. KingGeorge

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

21. KingGeorge

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

22. anilorap

yea... but u are using the right one

23. anilorap

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

24. KingGeorge

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

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

26. KingGeorge

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

ok let em check again

28. KingGeorge

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

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