Direct products - Is this an abuse of notation? 
Let $G=\langle x\rangle\times\langle y\rangle$ where $|x|=8$ and $|y|=4$. Find all pairs $a,b$ in $G$ s.t. $G=\langle a\rangle\times\langle b\rangle$, where $a$ and $b$ are expressed in terms of $x$ and $y$. (Abstract Algebra: Dummit & Foote, Direct products, Ex. 15)

Is this an abuse of notation?: here elements of $G$ are ordered pairs in the form of $(x^i,y^j)$. If $a,b\in G$, we can say that $a=(x^{i_1},y^{j_1})$ and $b=(x^{i_2},y^{j_2})$, so $G=\langle(x^{i_1},y^{j_1})\rangle\times\langle(x^{i_2},y^{j_2})\rangle$(!), with elements being ordered pairs of ordered pairs?
So I decided to treat elements like $(x^i,y^j)\in G$ as $x^iy^j$, just like the authors always do. So $G=\{1,x,x^2,\dots,y,xy,x^2y,...\}$. But then for $G=\langle a\rangle\times\langle b\rangle$, we have $\langle x\rangle=\langle a\rangle$ and $\langle y\rangle=\langle b\rangle$ (or not?), so $a=x, x^3, x^5,$ or $x^7$ and $b=y$ or $y^3$. Can I also take $a=xy$, for example, because with such use of shorthand notation, we may as well take $a=xy$ and $b=y$, not necessary that $\langle x\rangle=\langle a\rangle$ and $\langle y\rangle=\langle b\rangle$ (?)

You must just find all pairs of elements $a, b\in G$ such that $|a| \cdot |b| = |G|$ and $\langle a \rangle \cap \langle b \rangle = \{0\}$.
Since $|G|=32$ and $G$ obvioulsy does not contain elements of order greater than $8$, the only possibilities are $|a|=4, \, |b|=8$ and $|a|=8, \, |b|=4$.
Every element of $G$ is of the form $x^iy^j$, so you can now perform an easy case-by case analysis and complete the exercise. 
