The integers and the integers modulo a prime are both groups under addition.
What about the computer representation of integers (e.g. int64)?
It's closed under addition, since a sum that is too large wraps around to the negatives. It also inherits the other group properties from the integers (associativity, identity, inverse).
So int64 seems like a finite group, but am I missing anything?
Best Answer
If you just let overflows happen without doing anything about it, and in particular with 2's complement representation (or unsigned), a computer's $n$-bit integer representation becomes the integers modulo $2^n$. So yes, you're entirely right: It is a finite group (and with multiplication becomes a finite ring).
(As a side note, working with and thinking about 2's complement became a lot easier for me once I realized this. No one really told me during my education, so for ages I was stuck having to remember all the details in the algorithm for taking negatives, i.e. actually taking the 2's complement. Now that I have the algebraic knowledge of what's actually going on, I can just deduce the algorithm on the fly whenever I need it.)
It's not entirely obvious to check explicitly that they satisfy, say, associativity when overflow is in the picture. It's easier to set up the obvious bijection with the integers modulo $2^n$ and show that addition stays the same, and prove the group properties that way.