He didn't give it, but I attempted to come up with one myself. Here are the axioms
for vectors u,v,w in a vector space V and numbers a,b in a field K:
1) u + v = v + u
2) (u+v)+w = u+(v+w)
3) There is a zero vector O s.t v + O = v for all v
4) for every v and every O, there is an "opposite vector" v' s.t v + v' = O
5) a(bv) = (ab)v for all a,b, and v
6) (a+b)v = av + bv for all a,b, and v
7) a(u+v) = au + av for all a, u, and v
8) 1v = v for the number 1 in K
He says in a footnote that another mathematician informed him that (1) can be derived from (2) thru (8).
Here is an outline of what I did:
First, the zero vector is unique. Let O and O' be two zero vectors. Then O+O = O. But by (4), there is a vector "a" such that O + a = O'. Then
O + O + a = O + a
O + O' = O'
O = O' since O' is assumed to be a zero vector
The next useful fact is that 0v = O for all v. From (8) we have 1v = v, so
(0+1)v = v
0v + 1v = v
0v + v = v
0v + v + v' = v + v'
0v + O = O
0v = O
This also implies that for all v, O + v = v.
And then we need that for any v, its opposite v' is unique and is equal to -1v. From the preceding,
0v = O
(-1 + 1)v = O
-1v + 1v = O
-1v + v = O
-1v + v + v' = O + v' = v'
-1v + O = v'
-1v = v'
Now for any u,v, consider the vector w = u + v. Its opposite is w' = -1(u+v). But the vector w" = -1(v + u) is also an opposite for w since
w + w" = (u + v) + -1(v + u)
= (u+v) + -1v + -1u
= (u+v) + v' + u'
= u + (v + v') + u'
= u + O + u'
= u + u' = O
Since opposites are unique, -1(u + v) = -1(v + u), and multiplying by -1 gives
u+v = v+u.
