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Let $G$ be a linear algebraic group over a field $K$. (Say $K=\mathbb{F}_q$ or $K=\mathbb{C}$; do not assume $K$ is algebraically closed or of characteristic $0$.) Let $H_1$, $H_2$ be algebraic subgroups of $G$. Consider the multiplication map

$\phi:H_1\times H_2\to G$.

The image of $\phi$ is a constructible set, i.e., a variety $H$ with perhaps a few varieties of lower dimension deleted from it. (This is a special case of a result of Chevalley's.)

Question: when is $H_1(K) H_2(K)$ equal to $H(K)$?

There are two issues here: closure (i.e., really getting a variety rather than a constructible set as the image) and rationality.

Getting more specific, since the question above may be too hairy in general:

(a) Assume that $G$ is solvable. Does that help? Can we then answer the question in the affirmative?

(b) Say, furthermore, that both $H_1$ and $H_2$ are in the same unipotent subgroup of $G$, or that $H_1$ is unipotent and $H_2$ is a subgroup of a corresponding maximal torus. Does that help?

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Apologies for the trivial comment, but... There's another group-theoretical issue, surely: in general you wouldn't expect the image to be a subgroup unless $H_1$ and $H_2$ are commuting subgroups. – James Cranch May 13 '11 at 11:34
You are right of course - I mistyped. Still, it would be interesting to know whether one can say anything more if one knows H_1 H_2 to be a group. – H A Helfgott May 13 '11 at 13:03
A pedantic remark. If $K$ is a finite field then, as finite groups are algebraic groups for trivial reasons, when your set is a group, it is automatically an algebraic group. It's probably not the group you want though. – Felipe Voloch May 13 '11 at 14:17
Yes, I should really change the phrasing of my question to ask what I wanted to ask (and thereby break Paul Ziegler's answer below...) – H A Helfgott May 13 '11 at 16:55
Harald: I think it's a bit dangerous to think of a constructible set as "a variety with perhaps a few varieties of lower dimension deleted from it". Take for example the affine plane and then remove the x-axis and then re-insert the origin. That's constructible, but in some sense near the origin it is very far from being a variety: there is no sensible notion of an algebraic function in a neighbourhood of the origin as far as I know. In particular it seems to me to be "worse" then a variety with some bits missing---it's a variety with a bit added in quite a strange way. – Kevin Buzzard May 14 '11 at 8:28

The following is an answer to a previous version of the question, which asked whether there exists an algebraic subgroup $H$ of $G$ such that $H(K)=H_1(K)H_2(K)$:

There are two necessary conditions on $H_1, H_2$:

First, the set $\Gamma:=H_1(K)H_2(K)$ has to be a subgroup of $G(K)$.

Also, since any algebraic subgroup of an algebraic group over a field is closed, the set $\Gamma$ has to closed in $G$.

These conditions are also sufficient. This follows from from the following fact: Let $K$ be a field and $\Gamma$ a subgroup of $\operatorname{GL}_n(K)$ which is closed (for the Zariski toplogy on $\operatorname{GL}_n(K)$). Then there exists an algebraic subgroup $G$ of $\operatorname{GL}_n$ such that $G(K)=\Gamma$. This is (part of) Theorem 4.8 of these notes of Milne:

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Nice, but I just broke your answer by changing the question. Sorry. – H A Helfgott May 13 '11 at 16:55

This is not always true. Example: Let $G = \mathbb{G}_m^2$ with $H_1$ the subtorus $\{ (x,x) \in \mathbb{G}_m^2 \}$ and $H_2$ the subtorus $\{ (y,y^{-1}) \in \mathbb{G}_m^2 \}$. Note that $H_1 H_2 = G$. However, given $(u,v) \in G(K)$, we can write $(u,v)$ as $(xy, x y^{-1})$ if and only if $uv$ is square. So, if there are nonsquare elements of $K$, then $H_1(K) H_2(K) \neq G(K)$.

Here is a general sort of approach to these questions.

Let $F = H_1 \cap H_2$. Let $Y = H_1 H_2$. (Note that these are intersections and products of varieties.) Let $m$ be the map $H_1 \times H_2 \to Y$. For $x \in Y(K)$, the fiber $m^{-1}(x)$ is a torsor for $F$; namely, let $F$ act on $m^{-1}(x)$ by $(h_1, h_2) \mapsto (h_1 f^{-1}, f h_2)$.

So, if $H^1(\mathrm{Gal}(K), F)$ vanishes, then all torsors for $F$ over $K$ are trivial, and $m^{-1}(x)$ has a point, which is what you wanted.

For example, if the ambient group $G$ is unipotent, and we are in characteristic zero, then $F$ will automatically be unipotent. In particular, $F$ will have a filtration by $\mathbb{G}_a$'s and, as $H^1(\mathrm{Gal}(K), \mathbb{G}_a)=0$, we see that the statement is true in this case.

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What if $G$ is unipotent but we are not in characteristic zero? – H A Helfgott May 14 '11 at 13:37
Also, wait - what happens if $G$ is not unipotent but $F=\{e\}$? Would we automatically get that $H_1(K)H_2(K) = H(K)$, where $H$ is the Zariski closure of $H_1 H_2$? – H A Helfgott May 14 '11 at 14:02
Let $G = \mathbb{G}_a^2$ over a field of characteristic $p$. Let $H_1 = \{ (x,x) \}$ and $H_2 = \{ (y,y^p) \}$. Then $H_1 H_2$ is all of $G$. The point $(0,t)$ is in $H_1(K) H_2(K)$ if and only if $t$ is of the form $x^p-x$. So, if $K$ has nontrivial Artin-Schrier extensions, the result does not hold in this case. – David Speyer May 14 '11 at 14:04
Indeed, I am arguing that if $H_1 \cap H_2 = \{ e \}$ (scheme theoretically) then $(H_1 H_2)(K) = H_1(K) H_2(K)$. – David Speyer May 14 '11 at 14:05
So it is not enough to assume that $(H_1\cap H_2)(K) = \{e\}$, but it is enough to assume that $(H_1\cap H_2)(\overline{K}) = \{e\}$? (Is the latter statement exactly what you meant by "scheme-theoretically" in this context?) (Note: don't assume the ambient group is unipotent or solvable.) – H A Helfgott May 15 '11 at 9:09

Maybe it's worth pointing out that the question contained in the header has an obvious negative answer (and is not the main question being asked). The easiest counterexample would be the product of the two one-dimensional unipotent subgroups in the $3 \times 3$ upper triangular unipotent group which correspond to simple roots for the special linear group. This is a closed set (of dimension 2) but not a subgroup. On the other hand, the product of all three positive root groups in this situation is the entire upper triangular unipotent group. But here as in many natural solvable groups you are building a group step-by-step as product of a normal subgroup and another group.

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The following is a partial answer to the question whether $H_1 H_2$ is closed in $G$:

This is not true in general. For example, if $G$ is reductive, $T$ a maximal torus of $G$ and $B$ a Borel subgroup containing $T$, in the Bruhat decomposition of $G$ into double cosets $BwB$ for $w$ in the Weyl group with respect to $T$, for all but one $w$ the cell $BwB$ is not closed in $G$. For such a $w$ the product of the subgroups $B$ and $wBw^{-1}$ is not closed.

However, if $H_1$ and $H_2$ are unipotent, then $H_1 H_2$ is closed. This follows from the fact that any orbit under the action of a unipotent group on an affine variety is closed. This includes the first part of your (b).

$H_1 H_2$ is also closed in the situation of the second part of (b): Let $H_1$ be unipotent and $H_2$ a subgroup of a maximal torus $T$ which normalizes $H_1$ (I assume that's what you mean by corresponding maximal torus.) Since $H_1 H_2$ is a disjoint union of translates of $H_1^0 H_2$ we may assume that $H_1$ is connected. Then there exists a Borel subgroup $B$ of $G$ containing $T$ and $H_1$. Let $U$ be the unipotent radical of $B$. The product morphism $U\times T\to B$ is an isomorphism of varieties and under this isomorphism, $H_1 H_2$ corresponds to $H_1\times H_2$. Thus $H_1H_2$ is closed in $B$ and hence in $G$.

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Hi Paul - Thanks for your answer. This takes care of closure, but what about rationality? See the exchange between David Speyer and myself above. – H A Helfgott May 16 '11 at 13:35

If your subgroups are $\{exp(At)\}_t$ and $\{exp(Bt)\}_t$ and $[A,B]$, $A$, $B$ are linearly independent then your statement is false.

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