We are given a set G which is closed under the given binary operation(juxtaposition) such that the following axioms hold:
1.a(bc)=(ab)c for all a,b,c in G.
2.An element e exists in G such that ea=a for all a in G.Left identity
3.There exists a left inverse a' in G such that a'a=e for every element a in G.Left inverse
We need to prove that G satisfies the original group axioms.
Now,
ex=x for all x in G.
Therefore, ee=e. ......(1)
Now, a'a=e ......(2)
From (1) and (2),
(a'a)e=e ........(3)
From axiom 3 it follows that every element in G has a left inverse,which implies
a''a'=e for an a'' in G.
Pre-Multiplying b/s of (3) by a'' we have
a''(a'a)e=a''e
or,(a''a')ae=a''e(associative law)
or, ae=a''e (as a''a'=e) ........(4)
Also,a''e=a''(a'a)=(a''a')a=ea ........(5)
From (4) and (5)
ae=ea=a(from axiom 2) ........(6)
Now,ae=a''e
or,ea=ea'' (from (6))
or,a=a''
Thus,
e=a''a'=aa'
which together with a'a=e implies
a'a=aa'=e
Hence,G is a group under the given operation.
On the other hand the existence of a left identity and a right inverse seems to be insufficient because the above proof breaks down midway between equation 3 and 4.

I would do it this way: Assume from Penrose that:


Then we also have for any , so let's use (from Penrose's hint).

Then consider :

(with free use of the distributive law).

We then also have


Done.

As for the second part, I would have thought the only way really to do that would be to find a counter-example. Anyone?

For the second part, given that now , you would have to write:

or


Neither of which leads to as far as I can see!

Another way is to say:
Given that , and , show that and that .
So to prove that , let's ask what satisfies ?
So






To prove that let's say that since is, by definition, a member of the group it too must have an inverse, say .
Then from the definition of the inverse


So


since by the proof above.
So

However, if we start with and then we can't prove that and so, since the second line of the second proof above depends on this fact, we can't prove that either.

On the other hand, if we start with and then we CAN show that and using a similar proof to those above.

I have now posted this as an alternative solution.
6th October 2009

Here is my proposal of a counter-example:

Consider the set of mappings from a set onto itself, wherein are defined as follows:
(i.e. maps everything onto )
(i.e. maps everything onto )

Then is a "right identity" in the sense that

(proof "for all" : , i.e. , and , i.e. )

Furthermore, the definitions and provide proper "left inverses" in the sense that
.

However, is not a "left identity" because


The set therefore has a "right identity" and "left inverses" for each element, but the "right identity" is no "left identity".

jbeckmann, good example, but you can make it easier to understand.

Simply take a set S of (any number of) elements and define "multiplication" on the elements such that:

xy = x for all x,y in S.

Then clearly (xy)z = x(yz) = x so the associative law holds.
Any element will work as a right identity: xy=x for all x,y
so choose any particular element to be 'the' identity, 1.

then 1x = 1 for all x, so every element has the same left-inverse, and it's the identity, 1.

Clearly there is no left identity, nor right inverse, so associativity + right identity + left inverse is insufficient to guarantee these properties.