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

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

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
It 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.
 2 years ago

anilorap Group TitleBest ResponseYou've already chosen the best response.0
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

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

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

badreferences Group TitleBest ResponseYou've already chosen the best response.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.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.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.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.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.
 2 years ago

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

anilorap Group TitleBest ResponseYou've already chosen the best response.0
im still trying to understand the inverses
 2 years ago

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

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

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

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

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

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

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.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.
 2 years ago

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

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

anilorap Group TitleBest ResponseYou've already chosen the best response.0
ok let em check again
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.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.
 2 years ago

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