If $H$ is a normal subgroup of $G$, is $G/H \times H \cong G$? If $ H $ is a normal subgroup of $ G $, is $ G/H \times H \cong G $?
For example, I think $ \mathbb{Z}/2 \mathbb{Z} \times 2 \mathbb{Z} \cong \mathbb{Z} $. I would construct a map as follows:
\begin{align}
\phi: \mathbb{Z}/2 \mathbb{Z} \times 2 \mathbb{Z} &\longrightarrow \mathbb{Z}; \\
      (a,b)                                       &\longmapsto     a + b.
\end{align}
If it is not true in general, then are there a few criteria to show when it is true?
Thanks!
 A: No. If $A\times B\cong G$ then both $A$ and $B$ (respectively their images under the isomorphism) are normal in $G$. Your example with $\mathbb Z/2\mathbb Z$ does not work because $\mathbb Z/2\mathbb Z$ is not normal in $\mathbb Z$, yes it is not even a subgroup (no nonzero element $x\in \mathbb Z$ has the property $x+x=0$).
A: Any cyclic group whose order is not squarefree contains at least one normal subgroup for which this fails, as $C_{p^2m}\not\cong C_{p}\times C_{pm}$.
You may be looking for the Schur-Zassenhaus theorem.
A: Let me add a couple of comments to the other excellent answers. 
In some cases there is a subgroup $K$ of $G$ such that $G = H K$ and $H \cap K = 1$. Then $G/H \cong K$, and $G$ is a semidirect product (a.k.a. split extension) $H \rtimes K$ of $H$ by $K$. The semidirect product can be regarded as a generalization of the direct product, in which only one of the factors is normal.
In general, even such a subgroup $K$ of $G$ will not exist. Define then $K = G/H$. Now $G$ is said to be an extension of $H$ by $K$. Things get immediately very difficult here. However, one can define a section as a map $\sigma : K \to G$ such that $\varphi \circ \sigma = \textbf{1}_{K}$. (Here $\varphi : G \to K = G/H$ is the canonical map.) $\sigma(K)$ will not be a subgroup of $G$ (unless $G$ is a split extension of $H$ by $K$ as above), but the multiplication between elements of $\sigma(K)$ will have to be corrected by an element of $H$. This yields what is called a $1$-cocycle, and from now on you are in the territory of group cohomology.
A: Two trivial cases where this is true are $H = \{ 1 \}$ and $H = G$. Also, there exist many groups for which these are the only cases. One example is $S_3$, the nonabelian group of order $6$. If $S_3 \cong A \times B$, then $A = \{1\}$ or $A = S_3$.
A: It's worth noting that even when $G \cong H \times K$, in order for $G/H \times H$ to be isomorphic to $G$, you need to pick the "right" copy of $H$.
For example, let $G = C_2 \times C_4$, let $H = C_2$, and let $F = \{0\} \times \{0, 2\}$ be a subgroup of $G$.  Then $F \cong H$ and, since $G$ is abelian, $F$ is a normal subgroup.  However,
$$(G/F) \times H \cong C_2 \times C_2 \times C_2 \ncong G.$$
A: Supplementary Answer:
This fact is quoted from Serre's Course in Arithmetic(p.16 lemma).

Fact
  Let $A,C$ are finite Abelian groups with their orders are prime to each other.
  If we have exact sequence of Abelian groups
  $$
0\to A\to B\to C\to 0
$$
then $B\cong A\times C$.

Especially,let $G$ be a finite Abelian group and $H$ be a subgroup of $G$,we have 
$$
0\to H\to G\to G/H\to 0.
$$
So if gcd$(|H|,|G/H|)=1$, then $G\cong G/H\times H$.
