The Product of Subgroups of an Abelian Group Reference: Fraleigh p. 58 Question 5.43 in A First Course in Abstract Algebra 
Let $G$ be an Abelian Group. Suppose $H$ and $K$ are subgroups of $G$, and $HK = \{ xy: x \in H \text{ and } y \in K \}$. Prove that $HK$ is a subgroup of $G$.
Since $H$ and $K$ are both subgroups, $e \in H$ and $e \in K$. $ee \in HK$ which shows that $HK$ is not empty.
(i). Let $ab, cd \in HK$ such that $a, c \in H$ and $b, d \in K$. First I need to show that $abcd$ can be written as a product of elements of $H$ and elements of $K$. Since $G$ is Abelian, $abcd = acbd = (ac)(bd)$. We know that $(ac) \in H$ and $(bd) \in K$. 
(ii). Let $ab \in HK$, then $a \in H, b \in K$. I need to show that $(ab)^{-1} \in HK$ can be written as a product of elements of $H$ and $K$. Because $a \in H$ and $H$ is a subgroup, $a^{-1} \in H$. And since $b \in K$ and $K$ is a subgroup, $b^{-1} \in K$. $(ab)^{-1} = b^{-1}a^{-1}$.
Am I interpreting the question correctly? And is my proof correct?

@Arturo's Exercise:
I'm a bit confused as to the definition of $HK = KH$ given. I thought that if two groups $A$ and $B$ were equal, every element of $A$ would be contained in $B$ and every element of $B$ would be contained in $A$.
(i). Let $ab, cd \in HK$, then $abcd \in HK$. I need to show that $abcd \in HK$. Using the definition above, $ab = b'a'$ and $ab = a''b''$ where $a', a'' \in H$ and $b', b'' \in K$. A similar statement can be made for $cd$.
(ii). Let $ab \in HK$, then $(ab)^{-1} \in HK$
Could I get a hint as to how to start part (i)? I've tried substituting to try and show that $abcd$ is a product of elements of $H$ and $K$, but I have not gotten anywhere.
 A: The last part, as Geoff notes, is perhaps a bit lacking: you did not show that $b^{-1}a^{-1}$ is an element of $HK$, because that requires you to show that it can be written as the product of something in $H$ times something in $K$, rather than "something in $K$ times something in $H$" (yes, they are the same because $G$ is abelian, but this needs to be said somewhere).
Now, for further practice, you can try the usual problem:

Let $G$ be a group, not necessarily abelian, and let $H$ and $K$ be subgroups. Prove that $HK=\{ hk\mid h\in H,\ k\in K\}$ is a subgroup of $G$ if and only if $HK$ and $KH$ are equal as sets.

(Note that $HK=KH$ means that for each $h\in H$ and $k\in K$ there exist $h',h''\in H$ and $k',k''$ in $K$ such that $hk=k'h'$ and $kh = h''k''$. We do not require $hk=kh$ for each $h\in H$ and $k\in K$.)

Let $G$ be a group. If $A$ and $B$ are subsets of $G$, then we can form the subset $AB$,
$$AB = \{ab\mid a\in A, b\in B\}.$$
For example, take $G=S_3$, $A=\{(1,2), (1,3)\}$, $B=\{(1,2,3),(1,3,2)\}$. Then
$$AB = \{(1,2)(1,2,3), (1,2)(1,3,2), (1,3)(1,2,3), (1,3)(1,3,2)\} = \{(2,3), (1,3), (1,2)\}.$$
If $H$ and $K$ are subgroups of $G$, then $HK$ may or may not be a subgroup (it's certainly a subset). For instance, if $G=S_3$, $H=\{I,(1,2)\}$, $K=\{I,(1,3)\}$, then
$$\begin{align*}
HK &= \{II, I(1,3), (1,2)I, (1,2)(1,3)\}\\
 &= \{I, (1,3), (1,2), (1,3,2)\},
\end{align*}$$
which is not a subgroup, since it is not closed under products or inverses. On the other hand, if $H=\{I,(1,2)\}$ and $K=\{I, (1,2,3), (1,3,2)\}$, then
$$\begin{align*}
HK &= \{II, I(1,2,3), I(1,3,2), (1,2)I, (1,2)(1,2,3), (1,2)(1,3,2)\}\\
 &= \{I, (1,2,3), (1,32), (1,2), (2,3), (1,3)\}\end{align*}$$
which is a subgroup (in fact, it's the entire group).
Now, in the first example above, with $H=\{I, (1,2)\}$ and $K=\{I, (1,3)\}$, we can also construct $KH$. We have:
$$\begin{align*}
KH &= \{II, I(1,2), (1,3)I, (1,3)(1,2)\}\\
 &= \{I, (1,2), (1,3), (1,2,3)\}.\end{align*}$$
Notice that $KH$ is not equal to $HK$: $KH$ contains $(1,2,3)$, which is not in $HK$.
Now look at the second example above, with $K=\{I, (1,2,3), (1,3,2)\}$ and $H=\{I, (1,2)\}$. We have:
$$\begin{align*}
KH &= \{ II, I(1,2), (1,2,3)I, (1,2,3)(1,2), (1,3,2)I, (1,3,2)(1,2)\}\\
 &= \{I, (1,2), (1,2,3), (1,3), (1,3,2), (2,3)\}.\end{align*}$$
Here, $KH$ is equal to $HK$. However, also notice that it is not true that $hk=kh$ for each $h\in H$ and $k\in K$. $(1,3)$ is in both $HK$ and in $KH$, but $(1,3)$ appears in $HK$ as the product $(1,2)(1,3,2)$, whereas it appears in $KH$ as the product $(1,2,3)(1,2)$ (different $h$).
The proposition I suggest is asking you to prove that if the set $HK$ is equal to the set $KH$, then $HK$ is a subgroup; and conversely, that if $HK$ is a subgroup, then the set $HK$ must be equal to the set $KH$.
Now, what are you assuming in (i)? That the sets are equal, or that $HK$ is a subgroup?
If we assume that $HK$ is a subgroup, that means that (i) it contains $e$; (ii) it is closed under products; and (iii) it is closed under inverses. Your objective is to show that $HK=KH$. In order to show that $HK=KH$, you need to show that $HK\subseteq KH$ and that $KH\subseteq HK$.
So, in order to show that $HK\subseteq KH$, you need to show that if you have $x\in HK$, then $x\in KH$. So, let $x\in HK$. That means that there exist $h\in H$ and $k\in K$ such that $x=hk$. What you need to show is that $x\in KH$; that is, you need to show that there exists $k'\in K$ and $h'\in H$ (it's possible that $k=k'$, but then again it's possible that $k\neq k'$; we don't know!) such that $x=k'h'$.
Well... we know that $HK$ is a subgroup of $G$. Since $x\in HK$, then we must have $x^{-1}\in HK$; that means that there exist $h''\in H$ and $k''\in K$ such that $x^{-1}=h''k''$. Therefore...
Now finish this part of the argument, and then show that $KH\subseteq HK$.
Conversely, you need to show that if $HK=KH$, then $HK$ is a subgroup. You need to show that: (i) $e\in HK$ (easy, since $H$ and $K$ are both subgroups); (ii) that if $x,y\in HK$, then $xy\in HK$; and (iii) that if $x\in HK$, then $x^{-1}\in HK$.
Well, for (iii), for example: suppose that $x\in HK$. Since $x\in HK=KH$, then there exist $h\in H$ and $k\in K$ such that $x=kh$. What is $x^{-1}$? Is it in $HK$? Etc. Finish this off.
