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 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.
 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.

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KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2All you need to show, is that it also has a right identity, and a right inverse.

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0yea,,, I have hourssss trying to :(

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0That's true as well for nonAbelian groups, right?

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Wait, yes it is. I was derping; I forgot the exact definition of Abelian for a second.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2It is true for nonabelian groups too, but I'm starting to see why he was having problems. I keep getting that statements that use circular logic.

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0I know that in order to prove the right identity i use some how my left inverse... but i keep getting stuck

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Man, I'm too sleepy for this. :(

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0bad reference... please... i have 3 nights with no sleep :(

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0I'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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2According 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2As 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2That took some time to take apart and figure out exactly what was going on.

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0im still trying to understand the inverses

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2Any particular step in that proof?

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0well,, im trying to understand step by step... let me see if i get stuck

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Woops, I tried proving it backwards using the right inverse. I feel dumb.

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0if x is the inverse of a and if we have left inverse the will be (x*a) nooo (a*x)

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2If that was unclear, it means that \(x*a=e\).

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2And we want to show that \(a*x=e\) as well.

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0yea... but u are using the right one

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0yea but you can apply that statement like its true.. u want to get that statement

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2In 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.

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0but my understanding is that u cant just say x*a=e=a*x.. u have to get the answer

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2We 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 \(x*a=a*x=e\)

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2I 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.

anilorap
 3 years ago
Best ResponseYou've already chosen the best response.0ok thanks you very much I kind of getting it..
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