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I. Let $G$ be a nonempty set closed under an associative product, which in addition satisfies:

(a) There exists an $ein g$ such that $acdot e=a$ for all $ain G$.

(b) Give $ain G$, there exists an element $a^{-1}in G$ such that $acdot a^{-1}=e$.

Prove that $G$ must be a group under this product.

II. Prove, by an example, that right indentity element and left inverse does not imply that $G$ is group.

My solution:

I. Since $G$ is closed set under an associative product, i.e. if $a,b,cin G$ then $(acdot b)cdot c=acdot (bcdot c)in G$. Taking $c=e$ we get $(acdot b)cdot e=acdot (bcdot e)=acdot b in G$. We have shown that $cdot$ is binary operation. Since $ain G$ then $a^{-1}in G$ and we have the following identities $$a^{-1}=a^{-1}cdot e=a^{-1}cdot (acdot a^{-1})=(a^{-1}cdot a)cdot a^{-1}$$
Then $$e=a^{-1}cdot (a^{-1})^{-1}=((a^{-1}cdot a)cdot a^{-1})cdot (a^{-1})^{-1}=(a^{-1}cdot a)cdot (a^{-1}cdot (a^{-1})^{-1})=(a^{-1}cdot a)cdot e=a^{-1}cdot a$$
We have shown that $acdot a^{-1}=a^{-1}cdot a=e$
then $ecdot a=(acdot a^{-1})cdot a=acdot (a^{-1}cdot a)=acdot e=a$

We have shown that for this set $G$ the property of associativity, binary, identity and inverse hold $Rightarrow$ $G$ - group.

II. But II indeed is true. Lets take the set $G={a,b,e}$ and define the product $cdot$ by the following identities: $ecdot e=acdot e=bcdot e=e$ and $a^{-1}=b, b^{-1}=a$ and consider the following multiplication table for our set $G$

$begin{array}{c | c c c c c}
hlinehline
& e & a & b \
hline
e & e & b & b & \
a & a & a & e & \
b & b & e & a & \
hline
end{array}
$

It's easy to verify that conditions of second problem hold for our $G$, however, $G$ is not group since we can show that $b=a$.

Is my reasoning above correct?

EDIT: Maybe this is a duplicate but I would like to know if my solution is true since I have solved it by myself. Especiaaly I am interested in the solution of the second problem.

You don't need to verify that the operation is defined and associative: that's already given.

What you need to show is that

1. 
   $e$ is a left identity as well as a right identity (the latter condition is given)


2. 
   $a^{-1}a=e$, for every $ain G$
   

On the other hand, using $a^{-1}$ may be misleading, but your argument seems good. For the sake of clarity, I'll denote by $b$ and $c$ elements such that $ab=e$ and $bc=e$. Your argument becomes
$$
b=be=b(ab)=(ba)b
$$
then
$$
e=bc=((ba)b)c=(ba)(bc)=(ba)e=ba
$$
Therefore
$$
ea=(ab)a=a(ba)=ae=a
$$
Good work!

The operation you give the Cayley table of does not define a group structure on ${e,a,b}$, because $ea=eb$, but $ane b$ (this is better than saying that “we can show that $a=b$, which is false at the outset). So long as you verify it is associative, you have your counterexample.