Direct products - Is this an abuse of notation?

Question

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 $\langle y\rangle=\langle b\rangle$ (?)

Answer 1

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.

Answer 2

The thing is that the elements of the cyclic groups should be also treated as ordered pairs, when considered as elements of $G$. In particular $x := (x,e_y)$, where $e_y$ is the identity element of the second cyclic group.

And the notation $G=\langle(x^{i_1},y^{j_1})\rangle\times\langle(x^{i_2},y^{j_2})\rangle$ will not give you ordered pairs of ordered pairs as elements of $G$. It's a notation for the direct product of the two cyclic groups and the elements of the direct product are $(a,b)$, s.t. $a \in \langle x \rangle, b \in \langle y \rangle$ with binary operation $(a_1,b_1)*(a_2,b_2) = (a_1*a_2,b_1*b_2)$.

I guess your misunderstanding arises from the thought that the each factor in the direct product accounts for one "coordinate". This is not true, as each element consistes of as many "coordinates" as the number of factors in the direct product. So in your example you don't get two "coordinates" with two more "subcoordinates". In fact you get only two "coordinates" and the direct product tells you how the binary operation acts on those elements.

Using the notation $x^iy^j$ instead of $(x^i,y^j)$ is justified as the group $G$ is abelian, as direct product of cyclic groups.