Arsthbt

What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$?

Would the circle $S^1$ in $mathbb R^2$ be an example? what is $mathbb R^2/S^1$?

EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.

Every connected real abelian Lie group $G$ is isomorphic to $mathbbR^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrmrank(pi_1(G))$ and $m=dim G-n$.

Now, if you have a short exact sequence of abelian Lie groups

$$0to Hto Gto G/Hto 0$$

Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence

$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$

So, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired

EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that

$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$

and

$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$

The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:

$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$

(Below is for the non-abelian situation)
Here's a simple interesting example.

Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutative) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.

Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).