Does $+=\oplus$, $-x=x$, $identity=0$ form an additive group? Wanting to get a better feel for what makes an additive group, I am wondering if using the following definitions gives an additive group:

*

*$a + b := a \oplus b \forall a,b:\mathbb{N}$
where $\oplus$ is the bitwise XOR operator

*$-x := x$

*$identity := 0$
As I'm learning this rather informally, criticism of my notation would also be appreciated.
These definitions seem to satisfy the requirements I know of:

*

*$a+b=b+a := a \oplus b = b \oplus a$

*$a+0=a:=a\oplus 0 = a$

*$a+-a=0:=a\oplus a = 0$
 A: Yes, this is indeed an abelian$^*$ group.
$^*$This is the standard term for a group in which the group operation is commutative; that said, I have seen "additive" used for this before, as in your post.
You've already checked all but one of the axioms; the remaining axiom (which is quite important!) is associativity. EDIT: and closure, as Jyrki Lahtonen points out above. So you should verify that XOR is in fact associative. Note that some logical operations are not associative, such as $\rightarrow$, so this is genuinely nontrivial (although very easy to check).

Having confirmed your guess, let me now give a very lengthy but hopefully interesting and useful remark:
This group is a great example of an important group construction method - the direct sum. You've probably already seen each of the following:

*

*The group $\mathbb{Z}/2\mathbb{Z}$.


*The usual direct (or Cartesian) product of two or more groups.
Combining these two ingredients we can create a very big group indeed: the "$\mathbb{N}$-fold product" of $\mathbb{Z}/2\mathbb{Z}$, denoted $$\prod_\mathbb{N}\mathbb{Z}/2\mathbb{Z}.$$ Concretely, elements of this group are infinite binary sequences and the group operation is ... XOR!
Now the group in the OP can "fit into" this group in a very nice way: think about taking a given natural number's binary expansion and adding an endless trail of zeroes to the front. This gives an infinite binary sequence, and you can check that it in fact constitutes an injective group homomorphism from your group to really big product group $\prod_\mathbb{N}\mathbb{Z}/2\mathbb{Z}$.
But this homomorphism is not surjective: no natural number, for example, corresponds to the "all-$1$s" binary sequence. There's an easy way to describe the image of this homomorphism:

 It's the set of sequences with only finitely many $1$s.

This motivates the definition of the direct sum. Finite direct sums are the same as finite direct/Cartesian products. However, infinite direct sums are much smaller than infinite direct products: given a family of groups $G_i$ ($i\in I$), the direct sum $\bigoplus_{i\in I}G_i$ is the subgroup of the direct product $\prod_{i\in I}G_i$ consisting of all "$I$-indexed sequences" whose $i$th term is the identity of $G_i$ for all but finitely many $i$.
Early on the direct sum is rather mysterious, but down the road it will turn out to be very important. So I think it's worth pointing out that your group is (isomorphic to) an infinite direct sum of the two-element group.
