Unnecessary Abuse in Notation I am working on the following problem and finding the abuse of notation particularly irritating: 

Let $G$ be an (additive) abelian group with subgroups $H$ and $K$. Show that $G \simeq H \oplus K$ if and only if there are homomorphisms $\pi_1 : G \to H$, $\pi_2 : G \to K$, $i_1 : H \to G$ and $i_2 : K \to G$ such that $\pi_1 i_1 = 1_H$, $\pi_2 i_2 = I_K$, $\pi_1 i_2 = 0$, and $\pi_2 i_1 = 0$, where $0$ is the map sending every element onto the zero (identity) element, and $i_1 \pi_1(x) + i_2 \pi_2(x) = x$ for every $x \in G$. 

Clearly we are suppose to interpret $\pi_k$ as projections from direct products to factors, and $i_k$ as embeddings of groups into direct products of groups. But the author not only wants to say $G$ is isomorphic to $H \oplus K$ but just is $H \oplus K$; but at the same time $G$ isn't the direct product. This is typical of Hungerford and is really the only thing that irritates me about his book. How can we rewrite the problem without the unnecessary abuse in notation? The part giving me most trouble is "$i_1 \pi_1(x) + i_2 \pi_2(x) = x$ for every $x \in G$"
 A: Notice that, $i_k\pi_k:G\longrightarrow G$ because $\pi_k$ is applied before $i_k$. It follows that $i_k\pi_k(x)\in G$ for all $x\in G$. Since $+$ is the group operation of $G$, the last statement makes perfect sense and is not notational abuse
EDIT: Seeing your last comment, it seems you misunderstand the author's intent.
They clearly want you to think of $\pi_1$ as the projection from $H\oplus K$ onto $H$, but if the map were defined on $H\oplus K$, well, what would that tell you about $G$?
Instead, the author invites you to:
$\qquad(1)$: Assuming $G\simeq H\oplus K$, find a map $\pi_1$ (and its friends) defined on $G$ that has all properties you would expect of a a projection onto $H$.
$\qquad (2)$: Assuming there is a map $\pi_1$ (and its friends) defined on $G$ that has all properties you would expect of a a projection onto $H$, show that $G\simeq H\oplus K$.
A: Here's a proof of a stronger statement: Suppose $G,H,K$ are abelian groups. Then $G\simeq H\oplus K$ if and only if etc. Stronger because we don't assume that $H$ and $K$ are subgroups of $G$; that's actually irrelevant to the exercise as stated. Note that everywhere below I mean exactly what I say, and I prove exactly what we're asked to prove.
Suppose first that $G\simeq H\oplus K$. Let $f:G\to H\oplus K$ be an isomorphism. Define $P_1:H\oplus K\to H$ by $P_1(h,k)=h$; similarly define $P_2(h,k)=k$. Define $\pi_1:G\to H$ by $\pi_1=P_1f$. Similarly define $\pi_2=P_2f$. Define $i_1:H\to G$ by $i_1(h) = f^{-1}(h,0)$; similarly define $i_2(k)=f^{-1}(0,k)$.
Now suppose $x\in G$, and let $f(x) = (h,k)$. It follows that $\pi_1(x) =h$ and $\pi_2(x)=k$, so that $i_1\pi_1(x)+i_2\pi_2(x) = f^{-1}(h,0) + f^{-1}(0,k) = f^{-1}((h,0)+(0,k)) =  f^{-1}(h,k) = x$.
That's a proof of exactly what he said, or at least one direction. No abuse of notation, except for writing $f^{-1}(h,k)$ instead of $f^{-1}((h,k))$. And I left out the part about $\pi_ki_j$, which you can easily verify from the definitions. (From exactly the definitions I gave, not from some "intepretation" of them!)
Hmm. There's one part of the other direction that I don't quite see - maybe it will be clear if I write it all out. Suppose that $\pi_k$ and $i_k$ satisfy the given conditions, and define $f:G\to H\oplus K$ by $f(x)=(\pi_1(x),\pi_2(x))$. We need to show that $f$ is an isomorphism. 
It's clear that $f$ is a homomorphism. And $f$ is injective: If $f(x) = (0,0)$ then $\pi_k(x)=0$, so $x=i_1\pi_1(x)+i_2\pi_2(x)=0$. I don't see how to show $f$ is surjective.
No wait. Given $h\in H$ and $k\in K$, let $x=i_1(h) + i_2(k)$. Then those conditions on $\pi_ni_m$ show that $\pi_1(x) = h$ and $\pi_2(x)=k$, so $f(x)=(h,k)$. QED.
A: Not to beat a dead horse, but in fact $G$  need  not  be the internal direct sum of $H$ and $K$! Seems relevant to the question of whether we're "supposed to think of it as" an internal direct sum.
Note first that in my other answer I showed that a stronger result is true: Suppose $G$, $H$ and $K$ are abelian groups. Then $G\simeq H\oplus K$ if and only if there exist $\pi_k$ and $i_k$ such that etc.
Let $G=\mathbb Z\oplus\mathbb Z$. Let $H=\{(n,0)\}$ and $K=\{(0,2m)\}$. Then $G\simeq H\oplus K$, hence there exist $\pi_k$ and $i_k$ such that etc, although $G$ is certainly not the internal direct sum of $H$ and $K$.
Hence "$G$ is the internal direct sum if and only if etc" is simply false - good thing Hungerford never said that.
(Hmm, in that example it's true that $H\cap K=\{0\}$, so the internal sum $H+K$ is a direct sum, just not equal to $G$. Maybe more striking would be to take $H$ as above and $K=H$.)
