Is $G$ isomorphic to $\frac{G}{H}\times H$? Let $G$ be a group and $H$ a normal subgroup. Is $G$ necessarily isomorphic to $\frac{G}{H}\times H$?
 A: If $G$ is the cyclic group of order $4$, and $H$ is its two-element subgroup, then $H \times \frac{G}{H} \cong H \times H$ and has no element of order $4$, so cannot be isomorphic to $G$.
A: Not necessarily. $D_8$ has $\langle r \rangle \cong C_4$ as a subgroup where $r$ is an element of order $4$ and $D_8 / \langle r \rangle \cong C_2$, but $\left(D_8 / \langle r \rangle \right) \times \langle r \rangle \cong C_2 \times C_4$ is abelian while $D_8$ is not.
A: No.  The only normal subgroup of $S_3$, the symmetric group on $3$ elements, is $A_3 \cong \mathbb{Z}/3\mathbb{Z}$, the alternating group on three elements.  But 
$$  (S_3/A_3) \times A_3 \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/6\mathbb{Z} \not \cong S_3  \text{.}  $$
A: Not necessarily. For instance, $S_3$ has as a normal subgroup $A_3 \cong \mathbb{Z}/3\mathbb{Z}$ with quotient  $S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z}$. But, the product of these groups is abelian, while $S_3$ is not. 
A: Not only need they not be isomorphic, but $G$ need not even have a subgroup (much less a direct factor) isomorphic to $G/H$.  For example, $SL(2,5)$ has a center $Z$ of order 2, and $SL(2,5)/Z$ is isomorphic to $A_5$, but $SL(2,5)$ has no subgroup isomorphic to $A_5$.  
A: This only happens when $G$ is a direct product of two groups to begin with (and even in that situation not for all normal subgroups), i.e. when $G$ is not indecomposable. Some counterexamples:


*

*The symmetric group $S_n$ for $n\geq 3$

*The quaternion group $Q_8$

*Any semidirect product of indecomposable groups that is not direct

*Any cyclic group $\mathbb{Z}_{p^k}$ for prime $p$ and $k\geq 2$

*The integers $\mathbb{Z}$

*etc.


The situation you describe is actually quite rare. To see that note that for example this doesn't work even for groups that are products, e.g. $G=\mathbb{Z}_4\oplus\mathbb{Z}_2$. Now take the subgroup $H=\mathbb{Z}_2\oplus\mathbb{Z}_2$ and note that
$$H\oplus G/H\simeq \mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2\not\simeq G$$
