Is this a theorem? There exists a group homomorphism of a group $G$ into a group $H$ if and only if there exists a direct sum $G \approx \bigoplus_{i\in I} H_i$ that involves $H$, i.e. $H_i = H$ for some $i$.  
Proof:
$\Leftarrow$:  The projection homomorphism onto $H$. $\implies$:  Let $f:G\rightarrow H$ and let $K = \ker f$.  Then let's look at $H \times K$, with the group operation defined component-wise.  Define $\psi : G \rightarrow H \times K$ by fixing $K$ and mapping $G-K$ to $H$ by $f$, i.e. $\psi(x) = f(x)$ for $x \notin K$.  $\psi$ is a homomorphism.  Probably not. Though... I give up.  Is there a theorem like this out there?
 A: Here's an interesting consequence of this theorem:

Theorem: Every finite $p$-group is $(\mathbb{Z}/p\mathbb{Z})^n$ for some $n$ (i.e. elementary abelian).

Why? If $|G|=p^n$, then we know that $G$ has a subgroup of every order dividing $|G|$. In particular, there exists $N\leqslant G$ such that $|N|=p^{n-1}$. But, necessarily $N\unlhd G$ since $[G:N]$ is the smallest prime dividing $|G|$. But, $G/N\cong \mathbb{Z}/p\mathbb{Z}$. So, by your theorem, $G\cong (\mathbb{Z}/p\mathbb{Z})\oplus H$ for some group $H$ of order $p^{n-1}$. By continuing this process we get that, indeed, $G\cong (\mathbb{Z}/p\mathbb{Z})^n$.
This fact would sure have been helpful to know in my first algebra course :)
EDIT: Prompted by the comment of us2012, it seems prudent to note this theorem is patently false. Every group of order $p^n$ is cyclic if and only if $n=1$ (and in fact, every group of order $p^n$ is abelian if and only if $n=1,2$). This "theorem" was proven predicated on the incorrect claim of Enjoys Math.
A: By the isomorphism theorems, the image of a homomorphism $G\to H$ is isomorphic to some quotient of $G$, namely $G/K$ where $K=\ker$ is the kernel of the homomorphism. It suffices then to contemplate when an arbitrary quotient $G/K$ ($K\triangleleft G$ must be normal) is a direct summand.
We say $Q$ is a direct summand of $G$ if $Q\le G$ is a subgroup and $G\cong Q\times K$ for some $K\le G$; equivalently this means there is a $K\le G$ such that $G=QK$, $Q\cap K=1$ and $[Q,K]=1$ (i.e. all of $Q$ commutes with all of $K$ and vice-versa). Note $Q$ is a summand iff $K$ is a summand.
In general if $Q=G/K$ is a quotient of $G$, there is little expectation there will be a subgroup of $G$ isomorphic to this $Q$. And in general there is little expectation that, given a normal subgroup $K$ of the overgroup $G$, there is some complementary subgroup $Q$ for which $G\cong Q\times K$. Even if there is an embedding of the quotient $G/K=Q\hookrightarrow G$ back into $G$ itself such that $Q$ and $G$ have the same image under the projection $G\to G/K$, it is only enough to guarantee that $G=K\rtimes Q$ is a semidirect product. (Rather than $Q$ commuting with $K$, conjugation of $K$'s elements by $Q$'s elements generally acts as not-necessarily trivial automorphisms on $K$.)
Back to full generality, the sequence $K\to G\to G/K$ does not "split," i.e. the quotient $G/K$ admits no embedding in $G$ compatible with the projection $G\to G/K$. This is exactly why the group extension problem is nowhere near trivial as beginners might naively hope or suspect. For more background I suggest searching e.g. "group extension problem" and finding similar sources.
As Alex's answer highlights, even in the context of finite abelian group theory, which is relatively nice and tame, there is no quotient-as-direct-summand theorem. It would say for example that $Z_{p^2}\cong Z_p\oplus Z_p$, which is false: rather than sitting "next to" another subgroup, a subgroup could be "contained" within a larger one in an "irreducible" or "nonfactorizable" way.
