Are all groups algebraic?

Question

I know the definition of a group as a set with an operation that satisfies certain axioms. I have heard that there is something called an algebraic group and that this is a group with a topology such that the multiplication and inverses are continuous (or something like that).

My question is: Given a group $G$, can this group always be viewed as an algebraic group?

From the comments below, I am pretty sure that O don't mean a topological group. I am thinking about the algebraic variety stuff.

Answer 1

What you've described is a topological group. The definition of an algebraic group is somewhat more complicated. And algebraic groups aren't topological groups (although in certain contexts they give rise to topological groups). There isn't really a sense in which any group $G$ can be viewed as an algebraic group, unless you drop a finiteness condition usually included in the definition of an algebraic group. Every (abstract) group $G$ does give rise to something like an algebraic group (called a constant group scheme), but not knowing your background, I'm not sure this is something about which you'd want to know more.

EDIT: It's difficult to know what to say because I don't know the background of the OP, but I'll try for more detail. Algebraic groups over a field $k$ are, by definition (in a precise sense which I won't go into) related to finitely generated $k$-algebras, i.e., quotients of polynomial rings over $k$ in finitely many variables, and in some sense, requires only finitely many of these quotients to describe. If you take a finite group $G$ (just an abstract, finite group), then there is an algebraic group $\underline{G}$ over $k$ (so it has the right finiteness properties I've alluded to above) with the property that $\underline{G}(k)=G$ (the group of $k$-rational points is your original abstract group $G$), but note that in general an algebraic group is not determined by its group of $k$-rational points. Also, some people (not me) require algebraic groups to be connected, and $\underline{G}$ will not be connected if $G$ is non-trivial. In fact, as a scheme, it is a disjoint union of $\#G$ copies of $\mathrm{Spec}(k)$. We can do the same thing for an arbitrary (i.e. not necessarily finite) abstract group $G$, and obtain a $k$-group scheme locally of finite type $\underline{G}$ with the property that $\underline{G}(k)=G$, but this $G$ will have infinitely many connected components, and thus by most definitions doesn't deserve to be called an algebraic group over $k$. Better to call it simply a group scheme (more precisely, the constant group scheme over $k$ associated to the abstract group $G$).

So every abstract group yields a group scheme over $k$ whose group of $k$-rational points equal to the original abstract group, but only the finite groups yield what people might call algebraic groups. For infinite groups, you just get a $k$-group scheme.

Answer 2

I suppose you could always endow any group with the discrete topology (every subset is open). Then the group operations are trivially continuous.

Answer 3

What you describe is in fact known as a topological group, rather than an algebraic group. And yes, any group can be made topological by giving it the discrete topology (though this is usually not interesting.

An algebraic group, however, is slightly different (but of the same flavour). It is a group which is also an algebraic variety (usually assumed affine, in which case we call it a linear algebraic group). The multiplication and inversion are then required to be morphisms of varieties. But now it becomes less easy to see if any group can be algebraic in this sense (and in fact, if the group is too large compared to the ring over which one would like to make it algebraic, then it is not possible).

Answer 4

So you can regard any abstract group as a topological group by just giving it the discrete topology, but this usually defeats the purpose of topological groups, which is to exploit a meaningful topology on the group in order to do something.

Most of the answers here are saying something along the lines of:

"If you try to do the analogous stupid thing for algebraic groups, producing a group with a trivial and probably useless algebraic structure, there are some obstructions to this -- you can't do it in general."

While that's true, I doubt it's helpful to someone with such a basic question.

Algebraic groups, by and large, tend to be things that can be described as groups of matrices satisfying some straightforward conditions. So long as the conditions can be regarded as polynomials in the entries of the matrices somehow, you have an algebraic group, which allows you to use the machinery of algebraic geometry to study the group.

Of course if you have, say, a finite group, you can consider it as an algebraic group in a trivial way (and in fact this is important in some cases where you also want to use it in conjunction with some algebraic group with nontrivial structure), but this won't help you study the finite group in any way.