Proof Verification: Abstract Algebra May you tell me if my proof is correct? Thank you so much!
Here is the problem:

Proof by induction.
By the points 3 and 4, the left inverse of g, L, and the right inverse of g, R, exists. We need to prove that L=R for any finite set that follow 1) 2) 3) 4).
If g is the only element of G, as G is closed gg=g then g is the identity element e. Then L=g=R.
If a, g are the only elements of G, a and g are different. As G is closed aa=e and gg=e. Then L=R.
Suppose that G has n-1 elements, and suppose that for each of those different elements L=R. 
Now consider that G has n elements, all the n-1 elements of the previous set plus one extra element. Let call the n element x. Then by the points 3 and 4, Lx=xR=e.
If L=x then R=x and L=R. If L=R we are done. Then we need to suppose that L is not equal to R, and that L and R are one of the first n-1 elements that by the inductive step, all the elements have left and right inverse and they are equal.
Then Lx=xR=e implies ax=xb=e.
Also, ax=e implies x=$a^{-1}$
Additionally, xb= e implies x=$b^{-1}$
Then $a^{-1}$ = $b^{-1}$ implies a=b. Then L=R. 
Q.E.D.
For the problem 30. One example is the infinite set of the even numbers. The operation is the common multiplication. It is not a group for many reasons, one of them is that the identity element of multiplication, 1, is not a member of the set. However, premises 1) 2) 3) and 4) work on the set.
 A: Sorry, but your proof is incorrect under many respects.
To begin with, in order to prove existence of inverses you need to prove the existence of a neutral element, which is not stated in the hypotheses. Also, induction cannot be used, unless you prove the subset with $n-1$ elements is closed under the operation (which it isn't, because of Lagrange's theorem).
For $a\in G$, consider the maps $\lambda_a\colon G\to G$, $\lambda_a(x)=a*x$, and $\rho_a\colon G\to G$, $\rho_a(x)=x*a$.
By points 3 and 4, these maps are injective, so also surjective because $G$ is finite. Fix $a_0\in G$; by surjectivity of $\lambda_{a_0}$, there exists $r\in G$ such that $a_0*r=a_0$.
We want to show that $a*r=a$, for every $a\in G$. By surjectivity of $\rho_{a_0}$, there exists $a_1\in G$ with $a=a_1*a_0$, so
$$
a*r=(a_1*a_0)*r=a_1*(a_0*r)=a_1*a_0=a
$$
With a similar technique, we can prove there exists $l\in G$ with $l*a=a$, for every $a\in G$.
In particular, $r=l*r=l$, so we have found that $G$ has a neutral element $l=r=e$.
Now, given $a\in G$, there exists $a'\in G$ with $a*a'=e$ and also $a''\in G$ with $a'*a''=e$. Then
$$
a''=e*a''=(a*a')*a''=a*(a'*a'')=a*e=a
$$
and so $a'$ is the two-sided inverse of $a$.

The natural numbers under addition provide a simple counterexample for the infinite case.
A: "By the points 3 and 4, the left inverse of g, L, and the right inverse of g, R, exists"
I'd prefer you to be explicit about this. Why must there be $a^{-1}$ to pre-multiply $ab = ac$ by?
"If a, g are the only elements of G, a and g are different. As G is closed aa=e and gg=e"
Why must $aa = a$ and $gg=g$ both be simultaneously false? (This follows from the existence of left inverses, but I'm not satisfied that you've proved it.)
A: Besides what @PatrickStevens and @EthanBolker already pointed out in their great answers and comments, there's also a problem with using induction here. The first $n-1$ elements may not satisfy the four given conditions — they may not be closed under the $*$ operation (and in fact, it's very unlikely that they are). So you can't invoke the induction hypothesis here — if the conditions aren't satisfied, you can't draw the desired conclusion.
Finiteness here is a crucial condition, but not for induction. Here's a hint to get you started, which also shows why finiteness of $G$ is important here. For convenience of notation, let's say $n=|G|$ is the number of elements in $G$. Let's fix some element $a\in G$, and consider the set $\{a*x\mid x\in G\}$. This set has exactly $n$ elements in it, which all have to be distinct by property 3. So these $n$ elements are precisely the $n$ elements of $G$. (This is the part that may be violated in the infinite case.) In particular, one of them is $a$ itself, so there exists an element $e$ such that $a*e=a$.
Note that this $e$ isn't known to be the identity element yet! For now, we just know that it works as the identity only for $a$ and only from the right. But it certainly is the natural candidate to be the identity. So next you should show that also $e*a=a$ for this $a$, and then that $b*e=e*b=b$ for any $b\in G$. Only after you have your identity element are you ready to handle inverses.
