Given a group $H\unlhd G$, is it natural/accurate to say that $G \cong H \oplus (G / H)?$ Given a group $G$ and a normal subgroup $H \leq G$ , is it natural / accurate to say that
$$G \cong H \oplus (G / H)$$
This is what I feel is true intuitively, but I'm sure a group theorist could argue otherwise. 
 A: Let $G=\Bbb Z$ and $H=2\Bbb Z$. Then $H\lhd G$ and $$\begin{align}H\oplus(G/H)&\cong 2\Bbb Z\oplus (\Bbb Z/2\Bbb Z)\\
&\cong \Bbb Z\oplus\Bbb Z_2\\
&\not\cong\Bbb Z.
\end{align}$$
A: Take $G=\mathbb{Z_4}$. It has a normal subgroup isomorphic to $\mathbb{Z_2}$, and the quotient group is $\mathbb{Z_2}$ as well. But $\mathbb{Z_4}$ is not isomorphic to $\mathbb{Z_2}\times\mathbb{Z_2}$.
A: Let’s take a look at the sequence
$$1\to H \xrightarrow{i} G \xrightarrow{\pi} G/H \to 1$$
where $i$ is inclusion and $\pi$ is natural projection.
What you want to be true, that $G\cong H\oplus G/H$, is what it means to say that the sequence above splits. It is equivalent to the existence of a homomorphism $\phi: G\to H$ such that $\phi\circ i$ is the identity map on $H$. 

Let’s consider the sequence that Mark gives in his answer:
$$1\to \Bbb{Z}/2\Bbb{Z} \xrightarrow{i} \Bbb{Z}/4\Bbb{Z} \xrightarrow{\pi} \Bbb{Z}/2\Bbb{Z}\to 1.$$
There are two possible choices for $\phi$: either $\phi(1) = 1$ or $\phi(1)=0$. In the first case, we have
$$(\phi\circ i)(1) = \phi(2) = \phi(1) + \phi(1) = 0$$
and in the second case we have
$$(\phi\circ i)(1) = \phi(2) = 0.$$
Thus no such $\phi$ exists, so the sequence does not split. Clearly, in Mark’s example it is much easier to see that the sequence doesn’t split, but this sort of argument is common when talking about more complicated examples.

Follow up:
Note that in the above case, you are assuming that everything in $H$ and what you identify as the quotient $G/H$ within $G$ commute. This might not always be the case.
If you can find a map $\phi: G/H \to G$ so that $(\pi\circ \phi)$ is the identity on $G/H$, then it’s true that
$$G\cong H\rtimes G/H,$$
where the symbol denotes the semidirect product. For example, it’s true that $D_{2n}\cong \Bbb{Z}/n\Bbb{Z}\rtimes \Bbb{Z}/2\Bbb{Z}$. 
