Internal Direct Product Sums In $\Bbb Z$, let $H$ = $<5>$ and $K$ = $<7>$. Prove that $\Bbb Z = HK$. Does $\Bbb Z = H × K$?
Firstly, what is the difference between $HK$ and $H × K$? I think for both I have to show that there exists some linear combination of the two numbers such that they will span all of $\Bbb Z$, is that correct?
 A: The phrase "some linear combination of the two numbers such that they will span all of $\Bbb Z$" doesn't really mean anything. Perhaps you want to write "show every number is a linear combination of five and seven," that would be appropriate.
The notation $HK$ means $\{hk:h\in H,k\in K\}$. When we say $H\times K$ is an internal direct product, we are not only defining it to be $HK$, we are furthermore claiming every element of it is expressible in the form $hk$ ($h\in H,k\in K$) uniquely. The notation $HK$ does not imply any such uniqueness condition. (Also, the notation $H\times K$, interpreted as internal direct product, further implies that elements of $H$ commute with elements of $K$ and vice-versa, which $HK$ does not imply.)
An external direct product $A\times B$ is comprised of 2-tuples of the form $(a,b)$ with $a\in A,b\in B$; it is a construction out of two given groups $A$ and $B$, which do not have to have any relationship to each other (they aren't necessarily subgroups of a common group, for example). We are not talking about this construction in this problem.
(It really is unfortunate we are using multiplicative notations for an additive group...) To do the first part of the problem, you need to demonstrate that given any integer $n$ there is a way to write it as a linear combination $n=5a+7b$ for some integers $a,b$. It's enough to show this for $n=1$, do you see why? Next to see if $\Bbb Z$ is an internal direct product of $H$ and $K$, you need to figure out if every element of the form $5a+7b$ can be written that way for a unique pair of $a,b$; it will not be an internal direct product if you can say that $5a+7b=5c+7d$ but $(a,b)\ne(c,d)$ for some $a,b,c,d$.
A: $H\times K$ is the Cartesian product of $H$ and $K$.  In other words,
$$H\times K=\{(n,m):n\in H,m\in K\}.$$  
You are correct as far as showing $\Bbb Z=HK$.
