Do we have $(G/H)\times H \cong G$ for groups in general? After some thought I began to suspect $(G/H)\times H \cong G$, so I tried to construct an isomorphism by hand. I came up with $\varphi: (gH, h) \mapsto gh$ which came out to work provided $G$ is abelian. I can't think of anything if $G$ isn't abelian, but I'm pretty convinced it's true. How else can one prove isomorphism? (perhaps more importantly is this actually true; of course my difficulty would be easiest explained if this is false!)  
Edit: Outline of my 'proof':
$\varphi((gH,h))(fH,k)) = \varphi((gH,h))\varphi((fH,k)) = ghfk$ and $\varphi((gH,h)(fH,k)) = \varphi((gfH,hk)) = gfhk$. Hence $\varphi$ is multiplicative iff $G$ abelian.
Next, $(gH,h)^{-1} = (g^{-1}H, h^{-1})$, so $\varphi((gH,h)^{-1}) = \varphi((g^{-1}H, h^{-1})) = g^{-1}h^{-1} = (hg)^{-1} = (gh)^{-1} = \varphi((gH,h))^{-1}$. This again relied on $G$ being abelian, and shows that $\varphi$ is a homomorphism.
Now $\varphi(gH,h) = 1 \iff g=1$ and $h=1$; so $\ker(\varphi)=(H,1)$ and so $\varphi$ is injective.
Lastly, $\varphi(gH,1)\mapsto g$ is a surjection onto $G-H$ and $\varphi(H,h)\mapsto h$ is a surjection onto $H$. Therefore the image of $\varphi$ is $G-H \cup H = G$
This is why I believe that I have an isomorphism; I would be grateful if you could point out its errors!
 A: What you say you proved is not true 
( because your map is not well defined). Set $G=\mathbb{Z}_4$,$H=\{0,2\}$. $G\not\cong (G/H)\times H$. This is because $G$ has an element of order $4$, while $(G/H)\times H$ is actually isomorphic t0 $\mathbb{Z}_2\times \mathbb{Z}_2$ which does not have any element of order $4$.
Thus,  $(G/H)\times H \cong G$ is not even true for all abelian groups.
A: The statement is not true.
Consider $G=\mathbb{Z}$, $H=2\mathbb{Z}$. 
$\mathbb{Z}/\mathbb{2Z}\times\mathbb{2Z} \neq \mathbb{Z}$ since the group on the left has torsion (a nonzero element of finite order).  
A: The problem with your proof is that $\phi : G/H \times H \rightarrow G$ is not a well-defined map.  If you set $\phi(gH,h) = gh$, then you must check that if $gH = g'H$, then $\phi(gH,h) = \phi(g'H,h)$, i.e. $gh = g'h$.  But it is easy to see this can be false in general.
For example, using Seth's example, suppose $G = \mathbb{Z}$ and $H = 2 \mathbb{Z}$.  Then taking $g = 0, g' = 2$, $0 + H = 2 + H$, but (taking $h = 0$), $0 + 0 \neq 2 + 0$.
A: Here is an example with non-abelian groups. Let $G=S_3$ and let $H=A_3$, the latter being the odd permutations. Then $$G/H \times H \approx \mathbb{Z}/2 \times A_3 \approx \mathbb Z/2 \times \mathbb Z/3$$
But this cannot happen, as $S_3$ is not abelian.
