Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
Right identity and Right inverse implies a group

Let $(G,*)$ be a binary structure that has the following properties:

1) The binary operation $*$ is associative.

2) There exists an element $e \in G$ such that for all $a \in G$, $e*a=a$ (Existence of left Identity).

3) For all $a \in G$, there exists $b \in G$ such that $b*a=e$ (existence of left inverses)

Prove that $(G,*)$ is a group.

This is how I did it.

I want to check that


$a*(a'*a)=a$ , from here I cant go on, because I only have left inverses as well. So I know something is missing. And what should I do to find the right inverse?

share|cite|improve this question

marked as duplicate by azarel, mixedmath, Kannappan Sampath, lhf, Zev Chonoles Mar 21 '12 at 8:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have you looked at ? – scaaahu Mar 20 '12 at 4:35
Yea, but the truth is I don't understand what they are doing :( – anilorap Mar 20 '12 at 4:45
Did you check out both solutions given in that page? The solution by Martin Sleziak is pretty short and easy to understand. – Daniel Montealegre Mar 20 '12 at 5:02
@anilorap, once you understand the solution on that page, you will be able to figure out the answer to your own question. – scaaahu Mar 20 '12 at 5:29
anilorap, if I understand correctly, you had already seen the question about the right inverse when you posted yours? If so, it would have been much preferable to post a specific question with a link to the other question about what part of the answers there you have trouble understanding or transferring to the case of the left inverse, rather than to post a question that's very close to being an exact duplicate of the other one. (Two people have voted to close it as a duplicate.) – joriki Mar 20 '12 at 7:33
up vote 10 down vote accepted

Since it appears that you are very confused, I will lend you a hand. Take $a$ in your group $G$. Then by the conditions of the problem you know that it has a left inverse $b$. This means that you know that $$ba=e$$Where $e$ is the left identity. First we will show that $b$ is such that $ab=e$. This is how it goes: Let $y=ab$. Then $$yy=(ab)(ab)=a(ba)b=a(e)b=a(eb)=a(b)=ab=y$$Now since we know that we have left inverses, let us say that $q$ is the left inverse of $y$, so that $qy=e$. Multiply this at both sides and obtain: $$qyy=qy$$$$\Rightarrow ey=e$$$$\Rightarrow y=e$$$$\Rightarrow ab=e$$Hence by definition $b$ is then a $\textbf{right}$ inverse of $a$ also. Now by convention since you have seen that the right and left inverse of $a$ agree, people tend to call them $a^{-1}$ (I have not proved that it is unique, but you do not need to know this as of the moment).

Now let us prove that $e$ works as a right identity. $$ae=a(a^{-1}a)=(aa^{-1})a=ea=a$$ Above we use the fact that $a^{-1}a=e$, which we proved above, and the fact that $ea=a$ which is given as a condition of the problem.

share|cite|improve this answer
Thank you very much. I got stuck in the first part i had c*c=c.. after taht I didnt know what to do... Thank you very much :) – anilorap Mar 20 '12 at 7:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.