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!
On
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 $A = S_3$.
On
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.
On
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.
On
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, $$(G/F) \times H \cong C_2 \times C_2 \times C_2 \ncong G.$$
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 $x+x=0$).